近期的做题总结

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;scanf("%d",&T);
 6     while(T--)
 7     {
 8         int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);
 9         int t=(a==0)+(b==0)+(c==0)+(d==0);
10         if(t==4)puts("Typically Otaku");
11         if(t==3)puts("Eye-opener");
12         if(t==2)puts("Young Traveller");
13         if(t==1)puts("Excellent Traveller");
14         if(t==0)puts("Contemporary Xu Xiake");
15     }
16     return 0;
17 }

View Code

gym102028BUltraman vs. Aodzilla and Bodzilla

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=1e5+5;
 5 int ha,hb,ta,tb;ll ans[2];char s[2][N];
 6 int solab(ll &ans,char s[])
 7 {
 8     int i=0;ll sa=0;
 9     while(sa<ha){s[++i]='A';sa+=i;}
10     ans=(ll)i*ta;ll sb=sa-ha;
11     while(sb<hb){s[++i]='B';sb+=i;}
12     ans+=(ll)i*tb;ll x=(sa-ha)-(sb-hb);
13     if(x>0)s[sa-ha]='B';return i;
14 }
15 int solba(ll &ans,char s[])
16 {
17     int i=0;ll sb=0;
18     while(sb<hb){s[++i]='B';sb+=i;}
19     ans=(ll)i*tb;ll sa=sb-hb;
20     while(sa<ha){s[++i]='A';sa+=i;}
21     ans+=(ll)i*ta;ll x=(sb-hb)-(sa-ha);
22     int j=1;ll t=sb-hb;
23     while(t>=j){s[j]='A';x-=j;t-=j;j++;}
24     while(x>0){s[j-1]='B';s[j]='A';x--;j++;}
25     return i;
26 }
27 int main()
28 {
29     int T;scanf("%d",&T);
30     while(T--)
31     {
32         scanf("%d%d%d%d",&ha,&hb,&ta,&tb);
33         int p,n=solab(ans[0],s[0]);solba(ans[1],s[1]);
34         if(ans[0]!=ans[1])p=ans[0]<ans[1];
35         else{int i=1;while(i<n&&s[0][i]==s[1][i])++i;p=s[0][i]<s[1][i];}
36         p^=1;s[p][n+1]=0;printf("%lld %s
",ans[p],s[p]+1);
37     }
38     return 0;
39 }

View Code

gym102028CSupreme Command

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=300005;
 5 int T,n,m,vis[N];ll LL,LR,RL,RR,C[N];
 6 inline void cal(int i);
 7 struct D
 8 {
 9     int p[N],id[N];ll l,r,d;
10     void init(){l=1,r=n,d=0;}
11     void pre(){cal(id[1]);cal(id[n]);}
12     void stp(int x,int i){p[i]=x;id[x]=i;}
13     void left(int k){while(l<r&&l+d-k<1)cal(id[++l]);if(l+d-k>=1)d-=k;else d=1-l;}
14     void right(int k){while(r>l&&r+d+k>n)cal(id[--r]);if(r+d+k<=n)d+=k;else d=n-r;}
15     int qry(int x){return p[x]<=l?l+d:(p[x]>=r?r+d:p[x]+d);}
16 }A, B;
17 inline void cal(int i)
18 {
19     if(vis[i])return;
20     else if(A.p[i]<=A.l&&B.p[i]<=B.l)vis[i]=1,LL++;
21     else if(A.p[i]<=A.l&&B.p[i]>=B.r)vis[i]=1,LR++;
22     else if(A.p[i]>=A.r&&B.p[i]<=B.l)vis[i]=1,RL++;
23     else if(A.p[i]>=A.r&&B.p[i]>=B.r)vis[i]=1,RR++;
24 }
25 ll qry()
26 {
27     if(A.r-A.l>0&&B.r-B.l>0)return C[LL]+C[RR]+C[LR]+C[RL];
28     if(A.r-A.l==0&&B.r-B.l>0)return C[LL+RL]+C[LR+RR];
29     if(B.r-B.l==0&&A.r-A.l>0)return C[LL+LR]+C[RR+RL];
30     return C[LL+LR+RR+RL];
31 }
32 int main()
33 {
34     scanf("%d",&T);
35     for(int i=2;i<N;i++)C[i]=1ll*i*(i-1)/2;
36     while(T--)
37     {
38         scanf("%d%d",&n,&m);LL=LR=RR=RL=0;
39         for(int i=1,x,y;i<=n;i++)scanf("%d%d",&x,&y),B.stp(x,i),A.stp(y,i);
40         A.init();B.init();A.pre();B.pre();int k;char s[3];
41         for(int i=1;i<=m;i++)
42         {
43             scanf("%s",s);
44             if(s[0]=='!')printf("%lld
",qry());
45             else 
46             {
47                 scanf("%d",&k);
48                 if(s[0]=='L') A.left(k);
49                 else if(s[0]=='R') A.right(k);
50                 else if(s[0]=='U') B.left(k);
51                 else if(s[0]=='D') B.right(k);
52                 else printf("%d %d
",B.qry(k),A.qry(k));
53             }
54         }
55         for(int i=1;i<=n;i++)vis[i]=0;
56     }
57     return 0;
58 }

View Code

gym102028DKeiichi Tsuchiya the Drift King

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;scanf("%d",&T);
 6     while(T--)
 7     {
 8         double a,b,r,d;scanf("%lf%lf%lf%lf",&a,&b,&r,&d);
 9         double l=sqrt(b*b+(a+r)*(a+r)),ang=acos((a+r)/l);d=d/180*acos(-1.0);
10         if(d>ang||fabs(d-ang)<1e-6)printf("%.9lf
",l-r);
11         else{double l2=(a+r)/cos(d),x=(b-l2*sin(d))*sin(d);printf("%.9lf
",l2+x-r);}
12     }
13     return 0;
14 }

View Code

gym102028EResistors in Parallel

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 string S[205],s1[205],s2[205]; 
 4 int d[205],f[205][505];
 5 inline void prt(string s)
 6 {
 7     int t=0;while(s[t]=='0')t++;
 8     for(int i=t;i<=200;i++)printf("%c",s[i]);
 9 }
10 inline string mul(string a,int b){for(int j=200,c=0;j>=0;j--){int t=(a[j]-'0')*b+c;a[j]='0'+t%10;c=t/10;}return a;}
11 int main()
12 {
13     string s="";for(int i=1;i<=200;i++)s+="0";s+="1";
14     int cc;S[cc=1]=s1[1]=s2[1]=s;
15     for(int i=2;i<=500;i++)
16     {
17         int ff=1;for(int j=2;j<i;j++)if(i%j==0){ff=0;break;}
18         if(!ff)continue;
19         int t1=i,t2=i+1,c=0;
20         for(int j=200;j>=0;j--)
21         {
22             int t=(S[cc][j]-'0')*t1+c;
23             d[j]=t%10;c=t/10; 
24         }
25         if(c)break;
26         S[cc+1]="";
27         for(int j=0;j<=200;j++)S[cc+1]+=('0'+d[j]);
28         cc++;
29         for(int j=2;j<500;j++)f[cc][j]=f[cc-1][j];f[cc][i]++;
30         for(int j=2;j<500;j++)if(t2%j==0){while(t2%j==0)f[cc][j]--,t2/=j;}
31         s1[cc]=s2[cc]=s;
32         for(int j=2;j<500;j++)if(f[cc][j]>0)for(int k=1;k<=f[cc][j];k++)s1[cc]=mul(s1[cc],j);
33         for(int j=2;j<500;j++)if(f[cc][j]<0)for(int k=1;k<=-f[cc][j];k++)s2[cc]=mul(s2[cc],j);
34     }
35     int C;cin>>C; 
36     while(C--)
37     {
38         cin>>s;int n=s.size();while(n<201)s="0"+s,n++;
39         int t=upper_bound(S+1,S+cc+1,s)-S-1;
40         prt(s1[t]);putchar('/');prt(s2[t]);putchar('
');
41     }
42     return 0;
43 }

View Code

gym102028FHoneycomb

 1 #include<bits/stdc++.h>
 2 using namespace std; 
 3 const int N=1005;
 4 int r,c,w,h,vis[N*6][N*6],d[2][6]={{2,-2,1,1,-1,-1},{0,0,3,-3,3,-3}};
 5 char s[N*6][N*6];
 6 struct qq{int x,y,d;}st;
 7 inline bool chk(int x,int y){return x>0&&x<w&&y>0&&y<h;}
 8 inline int bfs()
 9 {
10     queue<qq>Q;qq u,v;u=st;vis[u.x][u.y]=1;Q.push(u);
11     while(!Q.empty())
12     {
13         v=Q.front();Q.pop();if(s[v.x][v.y]=='T')return v.d;
14         for(int i=0;i<6;i++)
15         {
16             int x=v.x+d[0][i],y=v.y+d[1][i],xx=x+d[0][i],yy=y+d[1][i];
17             if(s[x][y]==' '&&!vis[xx][yy]&&chk(xx,yy)){vis[xx][yy]=1;u.x=xx;u.y=yy;u.d=v.d+1;Q.push(u);}
18         }
19     }
20     return -1;
21 }
22 int main()
23 {
24     int T;scanf("%d",&T);
25     while(T--)
26     {
27         scanf("%d%d",&r,&c);getchar();w=4*r+3;h=6*c+3;
28         for(int i=0;i<w;i++)gets(s[i]);
29         for(int i=0;i<w;i++)
30         {
31             int l=strlen(s[i]);
32             for(int j=0;j<l;j++)if(s[i][j]=='S'){st.x=i,st.y=j,st.d=1;break;}
33             if(st.d)break;
34         }
35         int ans=bfs();
36         printf("%d
",ans);
37         for(int i=0;i<w;i++)
38         {
39             int l=strlen(s[i]);
40             for(int j=0;j<l;j++)s[i][j]=vis[i][j]=0;
41         }
42         st.x=0;st.y=0;st.d=r=c=w=h=0;
43     }
44     return 0;
45 }

View Code

gym102028GShortest Paths on Random Forests

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1000005,P=998244353;
 4 int qn[N],qm[N],fac[N],ifac[N],r[N],f[N],g[N],h[N],c[N],t[N],A[N],B[N],C[N];
 5 inline int pw(int a,int b){int r=1;for(;b;b>>=1,a=1ll*a*a%P)if(b&1)r=1ll*r*a%P;return r;}
 6 inline void ntt(int n,int*a,int f)
 7 {
 8     int l=1;while((1<<l)<n)l++;
 9     for(int i=1;i<n;i++)r[i]=r[i>>1]>>1|(i&1)<<(l-1);
10     for(int i=0;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
11     for(int i=2;i<=n;i<<=1)
12     {
13         int wn=pw(3,(P-1)/i);if(f==-1)wn=pw(wn,P-2);
14         for(int j=0;j<n;j+=i)for(int k=0,w=1;k<(i>>1);k++,w=1ll*w*wn%P){int u=a[j+k],v=1ll*a[j+k+(i>>1)]*w%P;a[j+k]=(u+v)%P;a[j+k+(i>>1)]=(u+P-v)%P;}
15     }
16     if(f==-1)for(int i=0,iv=pw(n,P-2);i<n;i++)a[i]=1ll*a[i]*iv%P;
17 }
18 void inv(int n,int*a,int*b)
19 {
20     if(n==1){b[0]=pw(a[0],P-2);b[1]=0;return;}
21     inv(n>>1,a,b);int m=n<<1;
22     for(int i=0;i<m;i++)A[i]=i<n?a[i]:0,b[i]=i<n?b[i]:0;
23     ntt(m,A,1);ntt(m,b,1);
24     for(int i=0;i<m;i++)b[i]=1ll*(P+2-1ll*A[i]*b[i]%P)*b[i]%P;
25     ntt(m,b,-1);for(int i=n;i<m;i++)b[i]=0;
26 }
27 void ln(int n,int*a,int*b)
28 {
29     inv(n,a,b);int m=n<<1;
30     for(int i=0;i<n-1;i++)B[i]=1ll*a[i+1]*(i+1)%P;B[n-1]=0;
31     for(int i=n;i<m;i++)b[i]=B[i]=0;ntt(m,b,1);ntt(m,B,1);
32     for(int i=0;i<m;i++)b[i]=1ll*b[i]*B[i]%P;
33     ntt(m,b,-1);for(int i=n-1;i;i--)b[i]=1ll*b[i-1]*pw(i,P-2)%P;b[0]=0;
34 }
35 void exp(int n,int*a,int*b)
36 {
37     if(n==1){b[0]=1;b[1]=0;return;}
38     exp(n>>1,a,b);int m=n<<1;ln(n,b,C);
39     for(int i=0;i<m;i++)C[i]=i<n?(a[i]+P-C[i])%P:0;C[0]++;
40     ntt(m,b,1);ntt(m,C,1);for(int i=0;i<m;i++)b[i]=1ll*b[i]*C[i]%P;
41     ntt(m,b,-1);for(int i=n;i<m;i++)b[i]=0;
42 }
43 void mul(int n,int*a,int*b,int*c)
44 {
45     for(int i=0;i<n;i++)A[i]=a[i],B[i]=b[i];
46     int m=n<<1;for(int i=n;i<m;i++)A[i]=B[i]=0;
47     ntt(m,A,1);ntt(m,B,1);
48     for(int i=0;i<m;i++)c[i]=1ll*A[i]*B[i]%P;
49     ntt(m,c,-1);for(int i=n;i<m;i++)c[i]=0;
50 }
51 int main()
52 {
53     int T,n=0,nn=1;scanf("%d",&T);
54     for(int i=1;i<=T;i++)scanf("%d%d",&qn[i],&qm[i]),n=max(n,qn[i]);
55     while(nn<=n)nn<<=1;nn<<=1;ifac[1]=fac[0]=ifac[0]=1;
56     for(int i=2;i<nn;i++)ifac[i]=1ll*ifac[P%i]*(P-P/i)%P;
57     for(int i=1;i<nn;i++)fac[i]=1ll*fac[i-1]*i%P,ifac[i]=1ll*ifac[i-1]*ifac[i]%P;
58     c[1]=1;for(int i=2;i<=n;i++)c[i]=1ll*pw(i,i-2)*ifac[i]%P;exp(nn>>1,c,f);
59     c[1]=0;for(int i=2;i<=n;i++)c[i]=1ll*(i-1)*pw(i,i-1)%P*ifac[2]%P*ifac[i]%P;
60     mul(nn>>1,f,c,g);for(int i=1;i<=n;i++)c[i]=1ll*pw(i,i)*ifac[i]%P;
61     mul(nn>>1,c,c,t);c[0]=ifac[2];mul(nn>>1,t,c,h);mul(nn>>1,h,f,h);
62     for(int i=1;i<=n;i++)f[i]=1ll*f[i]*fac[i]%P,g[i]=1ll*g[i]*fac[i]%P,h[i]=1ll*h[i]*fac[i]%P;
63     for(int i=1;i<=T;i++)
64     {
65         int n=qn[i],m=qm[i],ans=1ll*f[n]*n%P*(n-1)%P*ifac[2]%P;
66         ans=(ans+P-g[n])%P;ans=1ll*ans*m%P*m%P;ans=(ans+h[n])%P;
67         printf("%d
",1ll*ans*pw(f[n],P-2)%P);
68     }
69     return 0;
70 }

View Code

gym102028HCan You Solve the Harder Problem?

 1 #include<bits/stdc++.h>
 2 #include<unordered_map>
 3 using namespace std;
 4 #define ll long long
 5 const int N=2e5+5;
 6 int n,tot,lst,rt,a[N],lk[N<<1],len[N<<1],stk[N],ps[N<<1],f1[N][30];
 7 ll f2[N][30];
 8 unordered_map<int,int>ch[N<<1]; 
 9 inline void extend(int c,int x)
10 {
11     int p=lst,np=lst=++tot;len[np]=len[p]+1;ps[np]=x;ch[p].clear();
12     for(;!ch[p].count(c);p=lk[p])ch[p][c]=np;
13     if(!p)lk[np]=rt;
14     else
15     {
16         int q=ch[p][c];
17         if(len[q]==len[p]+1)lk[np]=q;
18         else
19         {
20             int nq=++tot;ps[nq]=ps[q];len[nq]=len[p]+1;lk[nq]=lk[q];lk[q]=lk[np]=nq;ch[nq]=ch[q];
21             for(;ch[p][c]==q;p=lk[p])ch[p][c]=nq;
22         }
23     }
24 }
25 long long sol()
26 {
27     int tp=0;stk[0]=0;a[0]=1e9+7;ll ans=0;
28     for(int i=1;i<=n;i++)f1[i][0]=f2[i][0]=0;
29     for(int i=n;i>=1;stk[++tp]=i--)while(a[stk[tp]]<a[i]){f1[stk[tp]][0]=i;f2[stk[tp]][0]=1ll*(stk[tp]-i)*a[stk[tp]];tp--;}
30     for(int i=1;i<=18;i++)for(int j=1;j<=n;j++)f1[j][i]=f1[f1[j][i-1]][i-1],f2[j][i]=f2[j][i-1]+f2[f1[j][i-1]][i-1];
31     for(int i=2;i<=tot;i++)
32     {
33         int t1=ps[i]-len[lk[i]]+1,x=ps[i];
34         for(int j=18;j>=0;j--)if(f1[x][j]>=t1)x=f1[x][j];
35         int t2=ps[i]-len[i]+1;ans+=1ll*(t1-max(f1[x][0]+1,t2))*a[x];
36         x=f1[x][0];if(x<t2)continue;
37         for(int j=18;j>=0;j--)if(f1[x][j]>=t2)ans+=f2[x][j],x=f1[x][j];
38         ans+=1ll*(x-t2+1)*a[x];
39     }
40     return ans;
41 }
42 int main()
43 {
44     int T;scanf("%d",&T);
45     while(T--)
46     {
47         tot=lst=rt=1;ch[1].clear();scanf("%d",&n);
48         for(int i=1;i<=n;i++)scanf("%d",&a[i]),extend(a[i],i);
49         printf("%lld
",sol());
50     }
51     return 0;
52 }

View Code

gym102028IDistance

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=100005;
 5 int n,a[N];ll s1[N],s2[N];
 6 int main()
 7 {
 8     int T;scanf("%d",&T);
 9     while(T--)
10     {
11         scanf("%d",&n);a[1]=0;
12         for(int i=2;i<=n;i++)scanf("%d",&a[i]),a[i]+=a[i-1];
13         s1[0]=0;for(int i=1;i<=n;i++)s1[i]=s1[i-1]+a[i];
14         s2[n+1]=0;for(int i=n;i>=1;i--)s2[i]=s2[i+1]+a[i];
15         int l=0,r=n+1;ll s=0;
16         for(int i=1;i<=n;i++)
17         {
18             int p;if(i&1)p=l+1;else p=r-1;
19             s=s+1ll*l*a[p]-s1[l]+s2[r]-1ll*(n-r+1)*a[p];
20             if(i&1)l++;else r--;printf("%lld",s);if(i<n)putchar(' '); 
21         }
22         puts("");
23     }
24     return 0;
25 }

View Code

gym102028JCarpets Removal

咕咕咕

gym102028KCounting Failures on a Trie

咕咕咕

gym102028LConnected Subgraphs

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=100005,P=1e9+7;
 5 int n,m,d[N],c[N],ps[N],id[N],w[N];
 6 vector<int>g[N],e[N];
 7 inline int pw(int a,int b){int r=1;for(;b;b>>=1,a=1ll*a*a%P)if(b&1)r=1ll*r*a%P;return r;}
 8 inline bool cmp(int x,int y){return d[x]<d[y];}
 9 inline ll f1111(){ll ans=0,iv=pw(24,P-2);for(int i=1;i<=n;i++){ll t=iv;for(int j=0;j<4;j++)t=t*(d[i]-j)%P;ans=(ans+t)%P;}return ans;}
10 inline ll f112(){ll ans=0;for(int x=1;x<=n;x++)for(int i=0,y;i<e[x].size();i++)y=e[x][i],ans=(ans+1ll*(d[x]-1)*(d[x]-2)/2%P*(d[y]-1))%P;return ans;}
11 inline ll f22(){ll ans=0;for(int x=1;x<=n;x++)for(int i=0,y,p=0;i<e[x].size();i++)y=e[x][i],ans=(ans+1ll*p*(d[y]-1))%P,p+=d[y]-1;return ans;}
12 inline int c3()
13 {
14     int ans=0;for(int i=1;i<=n;i++)w[i]=0;
15     for(int x=1;x<=n;x++)
16     {
17         for(int i=0;i<g[x].size();i++)w[g[x][i]]=1;
18         for(int i=0;i<g[x].size();i++)for(int j=0,y=g[x][i],z;j<g[y].size();j++)if(w[z=g[y][j]])ans=(ans+1+e[x].size()+e[y].size()+e[z].size()-6)%P;
19         for(int i=0;i<g[x].size();i++)w[g[x][i]]=0;
20     }
21     return ans;
22 }
23 inline int c4()
24 {
25     int ans=0;for(int i=1;i<=n;i++)w[i]=0;
26     for(int x=1;x<=n;x++)
27     {
28         for(int i=0;i<e[x].size();i++)for(int j=0,y=e[x][i],z;j<g[y].size();j++)if(ps[x]<ps[z=g[y][j]]){ans=(ans+w[z])%P;w[z]++;}
29         for(int i=0;i<e[x].size();i++)for(int j=0,y=e[x][i];j<g[y].size();j++)w[g[y][j]]=0;
30     }
31     return ans;
32 }
33 int main()
34 {
35     int T;scanf("%d",&T);
36     while(T--)
37     {
38         scanf("%d%d",&n,&m);
39         for(int i=1;i<=n;i++)d[i]=0,id[i]=i,e[i].clear(),g[i].clear();
40         for(int i=1,u,v;i<=m;i++){scanf("%d%d",&u,&v);d[u]++;d[v]++;e[u].push_back(v);e[v].push_back(u);}
41         sort(id+1,id+n+1,cmp);
42         for(int i=1;i<=n;i++)ps[id[i]]=i;
43         for(int x=1;x<=n;x++)for(int i=0,y;i<e[x].size();i++)if(ps[x]<ps[y=e[x][i]])g[x].push_back(y);
44         printf("%d
",(1ll*(f1111()+f112()+f22()-3ll*c3()-3ll*c4())%P+P)%P);
45     }
46     return 0; 
47 }

View Code

gym102056AExotic … Ancient City

利用$wleq30$可以模拟Kruskal的过程，每次加入所有边权为w的边，考虑新加入的边数=新减少的连通块数，转化为求解边权$leq w$的连通块数，然后差分。(意义:相当于抹去了边权的影响，Kruskal求MST转化成并查集求连通块数，从而加入边的顺序可以重新定义。)

考虑将$leq w$的边从左到右一列一列依次加入，维护当前2列的连通性(因为只关心加入边两端是否已经连通)，并查集合并时列小的挂列大的下面，考虑如何从1~i-1列的并查集得到1~i列的并查集，显然1~i列的并查集可以认为是1~i-1和2~i两个一样的并查集重叠，考虑在1~i-1(等价于2~i)列的答案上加边：1~i-1的并查集信息相当于在i-1列内部有加边，考虑维护加入的边，显然每次要么加入，则2i个元素的并查集中连通块数减少1，否则这条边对之后的结果无影响，由于连通块数由2n变成1，复杂度$O(malpha(n))$。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pp pair<int,int>
 5 #define X first
 6 #define Y second
 7 const int N=200005,W=31;
 8 int n,m,k,fa[N];ll a[N],ans[N];
 9 vector<pp>e[W];
10 int fnd(int x){return fa[x]==x?x:fa[x]=fnd(fa[x]);}
11 int main()
12 {
13     scanf("%d%d%d",&n,&m,&k);
14     for(int i=1,u,v,w;i<=k;i++){scanf("%d%d%d",&u,&v,&w);e[w].push_back(make_pair(u,v));}
15     for(int w=1;w<=30;w++)
16     {
17         for(int i=1;i<=n*2;i++)fa[i]=i;
18         memset(a,0,sizeof(a));vector<pp>g;
19         for(int i=1;i<=w;i++)for(int j=0;j<e[i].size();j++)
20         {
21             int u=fnd(e[i][j].X),v=fnd(e[i][j].Y+n);
22             if(u==v)continue;if(u>v)swap(u,v);
23             a[1]++;fa[u]=v;
24             if(u>n&&v>n)g.push_back(make_pair(u-n,v-n));
25         }
26         for(int i=2;i<=m;i++)
27         {
28             vector<pp>gg;
29             for(int j=0;j<g.size();j++)
30             {
31                 int u=fnd(g[j].X),v=fnd(g[j].Y);
32                 if(u==v){a[i]--;continue;}
33                 if(u>v)swap(u,v);fa[u]=v;
34                 if(u>n&&v>n)gg.push_back(make_pair(u-n,v-n));
35             }
36             g=gg;
37         } 
38         for(int i=2;i<=m;i++)a[i]+=a[i-1];
39         for(int i=2;i<=m;i++)a[i]+=a[i-1];
40         for(int i=1;i<=m;i++)if(w<30)ans[i]-=a[i];else ans[i]+=w*a[i];
41     }
42     for(int i=1;i<=m;i++)printf("%lld
",ans[i]);
43     return 0;
44 }

View Code

gym102056BMysterious … Host

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=5050;
 5 ll n,P,dp[N][5];
 6 int main()
 7 {
 8     scanf("%lld%lld",&n,&P);dp[1][1]=1;
 9     for(int i=2;i<=n;i++)
10     {
11         for(int j=1;j<i;j++)for(int k=1;k<=4;k++)(dp[i][min(4,k+1)]+=dp[i-j][k]*dp[j][1])%=P;
12         dp[i][1]=(dp[i][2]+dp[i][3]+dp[i][4]*2)%P;
13     }
14     for(int i=1;i<=n;i++)printf("%lld
",dp[i][1]);
15     return 0;
16 }

View Code

gym102056CHeretical … Möbius

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int M=5e4+10,inf=1e9+7,lcm=901800900;
 5 ll ans=inf,a[210],u[M],p[6],g[6]={4,9,25,49,121,169};
 6 vector<ll>pr,v[6];bool vis[M];map<ll,ll>mp;
 7 inline ll mu(ll x)
 8 {
 9     if(x<M)return u[x];
10     if(mp.count(x))return mp[x];
11     for(int i=0;i<pr.size();i++)
12     {
13         if(pr[i]*pr[i]>x)break;
14         if(x%(pr[i]*pr[i])==0)return mp[x]=0;
15         if(x%pr[i]==0)x/=pr[i];
16     }
17     return mp[x]=1;
18 }
19 void exgcd(ll a,ll b,ll&x,ll&y){if(!b){x=1;y=0;}else{exgcd(b,a%b,y,x);y-=a/b*x;}}
20 inline ll crt(int n,ll*a,ll*m)
21 {
22     ll M=m[0],R=a[0],x,y;
23     for(int i=1;i<n;i++)
24     {
25         ll c=(a[i]-R%m[i]+m[i])%m[i];exgcd(M,m[i],x,y);
26         x=(x%m[i])*(c%m[i]);x=(x%m[i]+m[i])%m[i];
27         R=M*x+R;M*=m[i];R=(R%M+M)%M;
28     }
29     return R;
30 }
31 void calc()
32 {
33     for(ll i=crt(6,p,g);i+199<=1e9;i+=lcm)
34     {
35         if(i>=ans)break;int ff=1;
36         for(ll j=i;j<=i+199;j++)if(mu(j)!=a[j-i+1]){ff=0;break;}
37         if(ff){ans=min(ans,i);break;}
38     }
39 }
40 void dfs(int x)
41 {
42     if(x==6){calc();return;}
43     random_shuffle(v[x].begin(),v[x].end());
44     for(int j=0;j<v[x].size();j++){p[x]=v[x][j];dfs(x+1);}
45 }
46 int main()
47 {
48     for(int i=1;i<M;i++)u[i]=1;
49     for(int i=2;i<M;i++)
50     {
51         if(!vis[i]){pr.push_back(i);for(ll j=1ll*i*i;j<M;j+=1ll*i*i)u[j]=0;}
52         for(int j=0;j<pr.size();j++)
53         {
54             ll t=1ll*i*pr[j];
55             if(t<M){vis[t]=1;if(i%pr[j]==0)break;}else break;
56         }
57     }
58     int tot=0;
59     for(int i=0;i<10;i++)
60     {
61         char s[30];scanf("%s",s);
62         for(int j=1;j<=20;j++)a[i*20+j]=s[j-1]-'0',tot+=s[j-1]-'0';
63     }
64     if(tot<=70){puts("-1");return 0;}
65     for(int i=0;i<6;i++)for(int j=1;j<=g[i];j++)
66     {
67         int f=1;for(int k=j;k<=200&&f;k+=g[i])f=a[k]^1;
68         if(f)v[i].push_back((g[i]-j+1)%g[i]);
69     }
70     for(int i=0;i<6;i++)if(v[i].size()==0){puts("-1");return 0;}
71     dfs(0);if(ans!=inf){printf("%d
",ans);return 0;}else puts("-1");
72     return 0;
73 }

View Code

gym102056DDeja vu of … Go Players

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T;scanf("%d",&T);
 6     while(T--)
 7     {
 8         int n,m;scanf("%d%d",&n,&m);
 9         puts(n>m?"No":"Yes");
10         for(int i=0,x;i<n+m;i++)scanf("%d",&x);
11     }
12     return 0;
13 }

View Code

gym102056EImmortal … Universe

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gym102056FInterstellar … Fantasy

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 struct P{int x,y,z;};
 5 inline P operator+(const P&a,const P&b){return (P){a.x+b.x,a.y+b.y,a.z+b.z};}
 6 inline P operator-(const P&a,const P&b){return (P){a.x-b.x,a.y-b.y,a.z-b.z};}
 7 inline int operator*(const P&a,const P&b){return a.x*b.x+a.y*b.y+a.z*b.z;}
 8 inline double len(const P&a){return sqrt(a*a);}
 9 inline ll sqr(int x){return 1ll*x*x;}
10 int main()
11 {
12     int T;scanf("%d",&T);
13     while(T--)
14     {
15         P o,a,b;int r;scanf("%d%d%d%d%d%d%d%d%d%d",&o.x,&o.y,&o.z,&r,&a.x,&a.y,&a.z,&b.x,&b.y,&b.z);
16         if((a-b)*(a-b)==0||(o-a)*(b-a)<=0||(o-b)*(a-b)<=0)printf("%.9lf
",len(a-b));
17         else
18         {
19             double h=sqrt((o-a)*(o-a)-(double)sqr((o-a)*(b-a))/((a-b)*(a-b)));
20             if(r<=h)printf("%.9lf
",len(a-b));
21             else
22             {
23                 double la=sqrt((o-a)*(o-a)-r*r),lb=sqrt((o-b)*(o-b)-r*r),ang=atan2(sqrt((o-a)*(o-a)-h*h),h)+atan2(sqrt((o-b)*(o-b)-h*h),h)-atan2(la,r)-atan2(lb,r);
24                 printf("%.9lf
",ang*r+la+lb);
25             }
26         }
27     }
28     return 0;
29 }

View Code

gym102056GOmnipotent … Garland

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1000005;
 4 int n,m,k,b[N];char t[N];
 5 vector<int>g[N],ans[N];
 6 struct cmp
 7 {
 8     bool operator()(int x,int y)
 9     {
10         if((!g[x].size())!=(!g[y].size()))return !g[x].size();
11         if((!x)!=(!y))return !x;return g[x].size()<g[y].size();
12     }
13 };
14 inline void psh(int i,int l,int r){while(((++l)%=n)!=r)g[i].push_back(l);}
15 bool sol() 
16 {
17     scanf("%d%d%d%s",&n,&m,&k,t);int tp=0;
18     for(int i=0;i<n;i++)if(t[i]=='B')b[++tp]=i;
19     if(n%k||n/k<m||tp<2*m)return 0;
20     if(m==1)
21     {
22         for(int i=0;i<n;i++)if(t[i]=='B'&&t[(i+1)%n]=='B'){ans[1].resize(n);for(int j=0;j<n;j++)ans[1][j]=j;return 1;}
23         return 0;
24     }
25     int len=0;for(int i=1;i<=m;i++)len+=b[i*2]-b[i*2-1]-1;
26     if(len>n-2*m-len){int t=b[2*m];for(int i=2*m;i>=2;i--)b[i]=b[i-1];b[1]=t;}
27     for(int i=1;i<=m;i++)
28     {
29         ans[i].resize(2);ans[i][0]=b[i*2-1];ans[i][1]=b[i*2];
30         psh(i,b[i*2-1],b[i*2]);psh(0,b[i*2],b[i*2%(2*m)+1]);
31     }
32     priority_queue<int,vector<int>,cmp>Q;
33     for(int i=0;i<=m;i++)Q.push(i);
34     for(int i=1;i<=m;i++)
35     {
36         while(ans[i].size()%k)
37         {
38             int x=Q.top();Q.pop();
39             if(x==i)
40             {
41                 int y=Q.top();Q.pop();
42                 ans[i].push_back(g[y].back());g[y].pop_back();
43                 Q.push(y);
44             }
45             else ans[i].push_back(g[x].back()),g[x].pop_back();
46             Q.push(x);
47         }
48     }
49     for(int x=m;x>=0;x--)
50     {
51         int i=x%m+1;
52         for(int j=0;j<g[x].size();j++)ans[i].push_back(g[x][j]);
53         g[x].clear();
54         while(ans[i].size()%k)ans[i].push_back(g[0].back()),g[0].pop_back();
55     }
56     return 1;
57 }
58 int main()
59 {
60     int T;scanf("%d",&T);
61     while(T--)
62     {
63         if(!sol())puts("No");
64         else
65         {
66             puts("Yes");
67             for(int i=1;i<=m;i++)
68             {
69                 sort(ans[i].begin(),ans[i].end());
70                 printf("%d",(int)(ans[i].size()));
71                 for(int j=0;j<ans[i].size();j++)printf(" %d",ans[i][j]);
72                 puts("");
73             }
74         }
75     }
76     return 0;
77 }

View Code

gym102056HSaintly … Coins

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gym102056IMisunderstood … Missing

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=105,M=5505;
 5 int n;ll a[N],b[N],c[N],dp[2][N][M];
 6 int main()
 7 {
 8     int T;scanf("%d",&T);
 9     while(T--)
10     {
11         scanf("%d",&n);
12         for(int i=1;i<=n;i++)scanf("%lld%lld%lld",&a[i],&b[i],&c[i]);
13         for(int i=0;i<2;i++)for(int j=0;j<=n;j++)for(int k=0;k<=5050;k++)dp[i][j][k]=-1;
14         dp[n&1^1][0][0]=0;ll res=0;
15         for(int i=n;i>=1;i--)for(int j=0;j<=n;j++)for(int k=0;k<=5050;k++)if(~dp[i&1^1][j][k])
16         {
17             dp[i&1][j+1][k+i]=max(dp[i&1][j+1][k+i],dp[i&1^1][j][k]+a[i]);
18             dp[i&1][j][k]=max(dp[i&1][j][k],dp[i&1^1][j][k]+1ll*(k-j*i)*b[i]);
19             dp[i&1][j][k]=max(dp[i&1][j][k],dp[i&1^1][j][k]+1ll*j*c[i]);
20         }
21         for(int i=1;i<=n;i++)for(int j=1;j<=5050;j++)res=max(res,dp[1][i][j]);
22         printf("%lld
",res);
23     }
24     return 0;
25 }

View Code

gym102056JPhilosophical … Balance

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=200010;
 5 int n,cc,cnt=1,lst=1,fa[N<<1],len[N<<1],ch[N<<1][26],tg[N<<1],hd[N<<1];char s[N];double dp[N<<1];
 6 struct E{int v,nxt;}e[N<<1];
 7 inline void add(int u,int v){e[++cc]=(E){v,hd[u]};hd[u]=cc;}
 8 inline void extend(int c)
 9 {
10     int p=lst,np=lst=++cnt;tg[np]=1;len[np]=len[p]+1;
11     for(;!ch[p][c];p=fa[p])ch[p][c]=np;
12     if(!p)fa[np]=1;
13     else
14     {
15         int q=ch[p][c];
16         if(len[q]==len[p]+1)fa[np]=q;
17         else
18         {
19             int nq=++cnt;len[nq]=len[p]+1;tg[nq]=0;fa[nq]=fa[q];fa[q]=fa[np]=nq;memcpy(ch[nq],ch[q],sizeof(ch[q]));
20             for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
21         } 
22     }
23 }
24 void dfs(int x)
25 {
26     if(tg[x]){dp[x]=len[x];return;}double s=0;
27     for(int i=hd[x];i;i=e[i].nxt)dfs(e[i].v),s+=1.0/(dp[e[i].v]-len[x]);
28     dp[x]=1.0/s+len[x];
29 }
30 void sol()
31 {
32     scanf("%s",s+1);n=strlen(s+1);cc=0;lst=cnt=1;
33     for(int i=0;i<=2*n+5;i++)hd[i]=0,memset(ch[i],0,sizeof(ch[i]));
34     for(int i=n;i>=1;i--)extend(s[i]-'a');
35     for(int i=2;i<=cnt;i++)add(fa[i],i);
36     dfs(1);printf("%.10lf
",dp[1]);
37 } 
38 int main(){int T;scanf("%d",&T);while(T--)sol();return 0;}

View Code

gym102056KDesperate … Fire Survive

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gym102056LEventual … Journey

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=100005;
 4 int n,m,a[N],c[2],d[N],f[N];
 5 int main()
 6 {
 7     scanf("%d%d",&n,&m);
 8     for(int i=0;i<n;i++)scanf("%d",&a[i]),c[a[i]]++;
 9     for(int i=0,u,v;i<m;i++)
10     {
11         scanf("%d%d",&u,&v);u--;v--;
12         if(a[u]!=a[v])d[u]++,d[v]++;
13     }
14     for(int i=0;i<n;i++)if(d[i])f[a[i]]++;
15     for(int i=0;i<n;i++)
16     {
17         int ans=0;
18         if(!d[i])
19         {
20             ans+=c[a[i]]-1;
21             ans+=2*f[a[i]^1];
22             ans+=3*(c[a[i]^1]-f[a[i]^1]);
23         }
24         else
25         {
26             ans+=c[a[i]]-1;
27             ans+=d[i];
28             ans+=2*(c[a[i]^1]-d[i]);
29         }
30         printf("%d%c",ans,i<n-1?' ':'
');
31     }
32     return 0;
33 }

View Code

gym102012ARikka with Minimum Spanning Trees

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll unsigned long long
 4 const int N=200500,M=200500,P=1e9+7;
 5 int n,m,tot,f[N];ll k1,k2;
 6 struct E{ll u,v,w;}e[N];
 7 inline void add(ll u,ll v,ll w){e[tot++]=(E){u,v,w};}
 8 inline bool cmp(E a,E b){return a.w<b.w;}
 9 inline ll fnd(ll x){return f[x]==-1?x:f[x]=fnd(f[x]);}
10 inline ll gt(){ll k3=k1,k4=k2;k1=k4;k3^=k3<<23;k2=k3^k4^(k3>>17)^(k4>>26);return k2+k4;}
11 inline void gen()
12 {
13     scanf("%d%d%llu%llu",&n,&m,&k1,&k2);
14     for(int i=1;i<=m;i++){ll u=gt()%n+1,v=gt()%n+1,w=gt();add(u,v,w);add(v,u,w);}
15 }
16 ll sol()
17 {
18     memset(f,-1,sizeof(f));sort(e,e+tot,cmp);ll cnt=0,ans=0;
19     for(int i=0;i<tot;i++)
20     {
21         ll u=e[i].u,v=e[i].v,w=e[i].w,f1=fnd(u),f2=fnd(v);
22         if(f1!=f2){ans=(ans+w)%P;f[f1]=f2;cnt++;}
23         if(cnt==n-1)break;
24     }
25     if(cnt<n-1)return 0;else return ans%P;
26 }
27 int main()
28 {
29     int T;scanf("%d",&T);
30     while(T--){tot=0;gen();ll ans=sol();printf("%lld
",ans);}
31     return 0;
32 }

View Code

gym102012BRikka with Line Graphs

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gym102012CRikka with Consistency

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define db double
 4 const int N=55;
 5 int n,a[N],vis[N][N][N];db b[N][N],d[N][N][N];
 6 struct D{double d;int x,y,z;};
 7 bool operator<(D a,D b){return a.d>b.d;}
 8 priority_queue<D>Q;
 9 inline void upd(int x,int y,int z,db w){if(d[x][y][z]>w)d[x][y][z]=w;Q.push((D){w,x,y,z});}
10 int main()
11 {
12     int T;scanf("%d",&T);
13     for(int i=0;i<=50;i++)for(int j=0;j<=50;j++)if(i!=j)b[i][j]=sqrt((i-j)*(i-j)+1)/abs(i-j);
14     while(T--)
15     {
16         scanf("%d",&n);
17         for(int i=0;i<=n;i++)scanf("%d",&a[i]);
18         for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)for(int k=0;k<=50;k++)d[i][j][k]=1e9,vis[i][j][k]=0;
19         d[0][n][0]=0;Q.push((D){0,0,n,0});
20         while(!Q.empty())
21         {
22             int x=Q.top().x,y=Q.top().y,z=Q.top().z;Q.pop();
23             if(vis[x][y][z])continue;vis[x][y][z]=1;
24             if(x&&a[x]==z)upd(x-1,y,z,d[x][y][z]+(a[x-1]==a[x]));
25             if(x<n&&a[x+1]==z)upd(x+1,y,z,d[x][y][z]+(a[x+1]==a[x]));
26             if(y&&a[y]==z)upd(x,y-1,z,d[x][y][z]+(a[y-1]==a[y]));
27             if(y<n&&a[y+1]==z)upd(x,y+1,z,d[x][y][z]+(a[y+1]==a[y]));
28             if(x<n&&y<n)
29             {
30                 if(min(a[x+1],a[x])<=z-1&&max(a[x+1],a[x])>=z-1&&min(a[y+1],a[y])<=z-1&&max(a[y+1],a[y])>=z-1)upd(x,y,z-1,d[x][y][z]+b[a[x+1]][a[x]]+b[a[y+1]][a[y]]);
31                 if(min(a[x+1],a[x])<=z+1&&max(a[x+1],a[x])>=z+1&&min(a[y+1],a[y])<=z+1&&max(a[y+1],a[y])>=z+1)upd(x,y,z+1,d[x][y][z]+b[a[x+1]][a[x]]+b[a[y+1]][a[y]]);
32             }
33         }
34         printf("%.10lf
",d[n][0][0]);
35     }
36     return 0;
37 }

View Code

gym102012DRikka with Subsequences

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=250,P=1e9+7;
 4 int n,a[N],f[N][N],s[N][N];char c[N][N];
 5 int main()
 6 {
 7     int T;scanf("%d",&T);
 8     while(T--)
 9     {
10         scanf("%d",&n);
11         for(int i=0;i<n;i++)scanf("%d",&a[i]),a[i]--;
12         for(int i=0;i<n;i++)scanf("%s",c[i]);
13         for(int i=0;i<n;i++)for(int j=0;j<n;j++)f[i][j]=0;
14         int ans=0;
15         for(int i=0;i<n;i++)
16         {
17             for(int j=0;j<n;j++)for(int k=0;k<n;k++)
18             {
19                 s[j+1][k+1]=c[a[j]][a[i]]=='1'?f[j][k]:0;
20                 s[j+1][k+1]=(s[j+1][k+1]+s[j+1][k])%P;
21                 s[j+1][k+1]=(s[j+1][k+1]+s[j][k+1])%P;
22                 s[j+1][k+1]=(s[j+1][k+1]+P-s[j][k])%P;
23             }
24             for(int j=0;j<n;j++)for(int k=0;k<n;k++)if(a[i]==a[j]&&a[i]==a[k]){ans=(ans+s[j][k]+1)%P;f[j][k]=(f[j][k]+s[j][k]+1)%P;}
25         }
26         printf("%d
",ans);
27     }
28     return 0;
29 }

View Code

gym102012ERikka with Data Structures

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gym102012FRikka with Nice Counting Striking Back

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gym102012GRikka with Intersections of Paths

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=400005,P=1e9+7;
 4 int n,m,k,ans,fac[N],ifac[N],fa[N][22],dep[N],f1[N],f2[N];
 5 vector<int>g[N];
 6 inline int pw(int a,int b){int r=1;for(;b;b>>=1,a=1ll*a*a%P)if(b&1)r=1ll*r*a%P;return r;}
 7 inline int C(int a,int b){return a<b||a<0||b<0?0:1ll*fac[a]*ifac[b]%P*ifac[a-b]%P;}
 8 void dfs(int x,int p)
 9 {
10     fa[x][0]=p;dep[x]=dep[p]+1;
11     for(int i=0;i<g[x].size();i++)if(g[x][i]!=p){int y=g[x][i];dfs(y,x);}
12 }
13 void dfss(int x,int p)
14 {
15     for(int i=0;i<g[x].size();i++)if(g[x][i]!=p){int y=g[x][i];dfss(y,x);f1[x]+=f1[y];f2[x]+=f2[y];}
16     if(x!=1)ans=(ans+P-C(f1[x],k))%P;ans=(ans+C(f2[x],k))%P;
17 }
18 inline int lca(int u,int v)
19 {
20     if(dep[u]<dep[v])swap(u,v);int sub=dep[u]-dep[v];
21     for(int i=20;i>=0;i--)if(sub>>i&1)u=fa[u][i];
22     if(u==v)return u;
23     for(int i=20;i>=0;i--)if(fa[u][i]!=fa[v][i])u=fa[u][i],v=fa[v][i];
24     return fa[u][0];
25 }
26 int main()
27 {
28     for(int i=fac[0]=1;i<N;i++)fac[i]=1ll*fac[i-1]*i%P;
29     ifac[N-1]=pw(fac[N-1],P-2);
30     for(int i=N-1;i;i--)ifac[i-1]=1ll*ifac[i]*i%P;
31     int T;scanf("%d",&T);
32     while(T--)
33     {
34         scanf("%d%d%d",&n,&m,&k);ans=0;
35         for(int i=1;i<=n;i++)g[i].clear();
36         for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);g[u].push_back(v);g[v].push_back(u);}
37         dfs(1,0);for(int i=1;i<=n;i++)f1[i]=f2[i]=0;
38         for(int i=1;i<=20;i++)for(int j=1;j<=n;j++)fa[j][i]=fa[fa[j][i-1]][i-1];
39         for(int i=1;i<=m;i++)
40         {
41             int u,v;scanf("%d%d",&u,&v);int t=lca(u,v);
42             f1[u]++;f1[v]++;f1[t]-=2;f2[u]++;f2[v]++;f2[t]--;f2[fa[t][0]]--;
43         }
44         dfss(1,0);printf("%d
",ans);
45     }
46     return 0;
47 }

View Code

gym102012HRikka with A Long Colour Palette

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define X first
 4 #define Y second
 5 const int N=300005;
 6 int n,m,ans[N];
 7 pair<pair<int,int>,int>a[N];
 8 int main()
 9 {
10     int T;scanf("%d",&T);
11     while(T--)
12     {
13         scanf("%d%d",&n,&m);int len=0,nw=0;
14         for(int i=0;i<n;i++)scanf("%d%d",&a[i].X.X,&a[i].X.Y),a[i].Y=i;
15         if(n<m)
16         {
17             printf("%d
",0);
18             for(int i=0;i<n;i++)printf("%d%c",1,i==n-1?'
':' ');
19             continue;
20         }
21         sort(a,a+n);priority_queue<pair<int,int> >q;
22         for(int i=0;i<m;i++)q.push(make_pair(0,i));
23         for(int i=0;i<n;i++)
24         {
25             int l=a[i].X.X;pair<int,int>p=q.top();q.pop();
26             len+=max(min(l,-p.X)-nw,0);nw=l;
27             p.X=-max(-p.X,a[i].X.Y);ans[a[i].Y]=p.Y;q.push(p);
28         }
29         len+=max(-q.top().X-nw,0);printf("%d
",len);
30         for(int i=0;i<n;i++)printf("%d%c",ans[i]+1,i==n-1?'
':' ');
31     }
32     return 0;
33 }

View Code

gym102012IRikka with Sorting Networks

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=55;
 4 int n,m,p,ans,a[N],u[N],v[N];
 5 void dfs(int x,int t)
 6 {
 7     if(x==0){ans+=t;return;}
 8     if(a[u[x]]<a[v[x]]){dfs(x-1,t);swap(a[u[x]],a[v[x]]);dfs(x-1,t);swap(a[u[x]],a[v[x]]);}
 9 }
10 int main()
11 {
12     int T;scanf("%d",&T);
13     while(T--)
14     {
15         scanf("%d%d%d",&n,&m,&p);ans=0;
16         for(int i=1;i<=m;i++)scanf("%d%d",&u[i],&v[i]),u[i]--,v[i]--;
17         for(int i=0;i<n;i++)a[i]=i;dfs(m,1);
18         for(int i=0;i<n-1;i++){swap(a[i],a[i+1]);dfs(m,-1);swap(a[i],a[i+1]);}
19         for(int t=0;t<n;t++)for(int i=0;i<n;i++)if(i!=t)
20         {
21             int c=0;
22             for(int j=0;j<n;j++)if(j==i)a[j]=t;else{if(c==t)c++;a[j]=c++;}
23             dfs(m,1);
24         }
25         printf("%d
",ans);
26     }
27     return 0;
28 }

View Code

gym102012JRikka with An Unnamed Temple

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gym102012KRikka with Ants

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gym102012LRikka with Grid Graphs

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 int main()
 5 {
 6     int T,n,m;scanf("%d",&T);
 7     while(T--)
 8     {
 9         cin>>n>>m;vector<string>mp(2*n-1);getline(cin,mp[0]);
10         for(int i=0;i<2*n-1;i++){getline(cin,mp[i]);while((int)mp[i].size()<2*m-1)mp[i]+=' ';}
11         unordered_map<ll,ll>dp[2];
12         int p=0,q=1;dp[p][0]=1;
13         for(int x=0;x<n;x++)for(int y=0;y<m;y++)
14         {
15             swap(p,q);dp[p].clear();
16             int cu=(x>0&&mp[2*x-1][2*y]=='|'),cl=(y>0&&mp[2*x][2*y-1]=='-');
17             for(unordered_map<ll,ll>::iterator it=dp[q].begin();it!=dp[q].end();it++)
18             {
19                 ll st1=it->first;vector<int>g(m);
20                 for(int i=0,k=0;i<m;i++)for(int j=0;j<m;j++,k++)g[i]|=(st1>>k&1)<<j;
21                 for(int u=0;u<=cu;u++)for(int l=0;l<=cl;l++)
22                 {
23                     vector<int>h=g;h.push_back(0);
24                     h[m]|=u<<y,h[y]|=(cu-u)<<m;
25                     if(y)h[m]|=l<<(y-1),h[y-1]|=(cl-l)<<m;
26                     for(int k=0;k<=m;k++)for(int i=0;i<=m;i++)if(h[i]>>k&1)h[i]|=h[k];
27                     bool ff=1;for(int i=0;i<=m;i++)ff&=h[i]>>i&1^1;
28                     if(!ff)continue;ll st2=0;
29                     for(int i=0,k=0;i<m;i++)for(int j=0;j<m;j++,k++)st2|=(ll)(h[(i==y?m:i)]>>(j==y?m:j)&1)<<k;
30                     dp[p][st2]+=it->second;
31                 }
32             }
33         }
34         ll res=0;for(unordered_map<ll,ll>::iterator it=dp[p].begin();it!=dp[p].end();it++)res+=it->second;printf("%lld
",res);
35     }
36     return 0;
37 }

View Code

gym102012MRikka with Illuminations

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=2010;
 5 int T,n,m,L,R,ans,t[N],f[N],po[N],res[N];
 6 struct P
 7 {
 8     ll x,y;int i;
 9     bool operator<(const P &b)const{return x*b.y-y*b.x>0;}
10     P operator-(const P &b)const{return (P){x-b.x,y-b.y,i};}
11 }ps,p[N],s[N];
12 int main()
13 {
14     scanf("%d",&T);
15     while(T--)
16     {
17         int n,m;scanf("%d%d",&n,&m);ans=m+1;
18         memset(f,0,sizeof(f));memset(t,0,sizeof(t));
19         for(int i=1;i<=n;i++)scanf("%lld%lld",&s[i].x,&s[i].y),s[i].i=i;
20         for(int i=1;i<=m;i++)
21         {
22             ll x,y;scanf("%lld%lld",&x,&y);ps=(P){x,y,0};
23             for(int j=1;j<=n;j++)p[j]=s[j]-ps;
24             sort(p+1,p+n+1);L=p[n].i;R=p[1].i-1;
25             if(R<L)R+=n;if(R>f[L])f[L]=R,t[L]=i;
26             if(R<=n&&n+R>f[n+L])f[n+L]=n+R,t[n+L]=i;
27         }
28         for(int i=1;i<=2*n;i++)if(f[i-1]>f[i])f[i]=f[i-1],t[i]=t[i-1];
29         for(int i=1;i<=n;i++)
30         {
31             int ff=0,cc=0;
32             for(int j=i;j<n+i;j=f[j]+1)if(f[j]+1<=j){ff=1;break;}else po[++cc]=t[j];
33             if(!ff&&cc<ans){ans=cc;for(int j=1;j<=cc;j++)res[j]=po[j];}
34         }
35         if(ans==m+1)ans=-1;printf("%d
",ans);
36         if(ans>0){for(int i=1;i<ans;i++)printf("%d ",res[i]);printf("%d
",res[ans]);}
37     }
38     return 0;
39 }

View Code

gym101955ASockpuppets

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=100005,P=1e9+7;
 4 int n,m,tot=0,f[N][12][22],g[N][12][22];char c[N];
 5 struct st
 6 {
 7     int ch[26],o;
 8     void init(){o=0;memset(ch,0,sizeof(ch));} 
 9 }p[N];
10 inline void init(){tot=1;p[1].init();}
11 void ins(char*s,int o)
12 {
13     int l=strlen(s+1),x=1;
14     for(int i=1,c;i<=l;x=p[x].ch[c],i++)if(!p[x].ch[c=s[i]-'a'])p[p[x].ch[c]=++tot].init();
15     p[x].o=o;
16 }
17 void dfs(int x)
18 {
19     for(int i=0;i<=10;i++)for(int j=0;j<=20;j++)f[x][i][j]=g[x][i][j]=0;
20     f[x][0][0]=1;
21     for(int i=0;i<26;i++)
22     {
23         int v=p[x].ch[i];if(!v)continue;dfs(v);
24         for(int j=10;~j;j--)for(int k=20;~k;k--)if(f[x][j][k])for(int l=0;l<=10-j;l++)for(int t=0;t<=20-k;t++)g[x][j+l][k+t]=(g[x][j+l][k+t]+1ll*f[x][j][k]*f[v][l][t])%P;
25         for(int j=0;j<=10;j++)for(int k=0;k<=20;k++)f[x][j][k]=g[x][j][k],g[x][j][k]=0;
26     }
27     if(!p[x].o)return;
28     for(int i=0;i<=10;i++)for(int j=0;j<=20;j++)if(f[x][i][j])
29     {
30         if(p[x].o==1)
31         {
32             if(i+1<=10)g[x][i+1][j]=(g[x][i+1][j]+f[x][i][j])%P;
33             if(j-1>=0)g[x][i][j-1]=(g[x][i][j-1]+1ll*f[x][i][j]*j)%P;
34             if(i+2<=10)g[x][i+2][j]=(g[x][i+2][j]+1ll*f[x][i][j]*(P+1)/2)%P;
35             if(j-2>=0)g[x][i][j-2]=(g[x][i][j-2]+1ll*f[x][i][j]*j*(j-1)/2)%P;
36             if(i+1<=10&&j-1>=0)g[x][i+1][j-1]=(g[x][i+1][j-1]+1ll*f[x][i][j]*j)%P;
37         }
38         else
39         {
40             if(i-1>=0)g[x][i-1][j]=(g[x][i-1][j]+1ll*f[x][i][j]*i)%P;
41             if(j+1<=20)g[x][i][j+1]=(g[x][i][j+1]+f[x][i][j])%P;
42         }
43     }
44     for(int i=0;i<=10;i++)for(int j=0;j<=20;j++)f[x][i][j]=(f[x][i][j]+g[x][i][j])%P;
45 }
46 int main()
47 {
48     int T,Cs=0;scanf("%d",&T);
49     while(T--)
50     {
51         init();scanf("%d%d",&n,&m);
52         for(int i=1;i<=n;i++)scanf("%s",c+1),ins(c,1);
53         for(int i=1;i<=m;i++)scanf("%s",c+1),ins(c,2);
54         dfs(1);printf("Case #%d: %d
",++Cs,f[1][0][0]);
55     }
56     return 0;
57 }

View Code

gym101955BSequences Generator

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gym101955CInsertion Sort

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=55;
 5 ll fac[N];
 6 int main()
 7 {
 8     int T,Cs=0;scanf("%d",&T);
 9     while(T--)
10     {
11         ll n,k,q;scanf("%lld%lld%lld",&n,&k,&q);
12         for(int i=fac[0]=1;i<N;i++)fac[i]=fac[i-1]*i%q;
13         if(n<=k){printf("Case #%d: %lld
",++Cs,fac[n]);continue;}
14         ll ans=((n-1)*(n-k)%q+1)%q*fac[k]%q;
15         printf("Case #%d: %lld
",++Cs,ans);
16     }
17     return 0;
18 }

View Code

gym101955DDiameter of a Tree

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=1000005;
 5 int n,m,S,tt,cc,cnt,res,rt,s1,s2,vis[N<<1],hd[N],sz[N],v[N];ll ans[N];double P[N],Q[N];
 6 struct E{int v,nxt,a,b;}e[N];
 7 inline void add(int u,int v,int a,int b){e[++tt]=(E){v,hd[u],a,b};hd[u]=tt;vis[tt]=0;e[++tt]=(E){u,hd[v],a,b};hd[v]=tt;vis[tt]=0;}
 8 struct G{int u,v,a,b;}g[N];
 9 struct st{ll k,b;}p[N],q[N],r[N*50];
10 inline bool operator<(const st&a,const st&b){return a.k==b.k?a.b>b.b:a.k<b.k;}
11 inline bool chk(st&a,st&b,st&c){return 1.0*(c.b-b.b)/(b.k-c.k)<=1.0*(b.b-a.b)/(a.k-b.k);}
12 struct qq{ll w;int i;}W[N];
13 inline bool cmp(qq a,qq b){return a.w<b.w;}
14 void dfs(int x,int p)
15 {
16     for(int i=hd[x];i;i=e[i].nxt)if(e[i].v!=p)dfs(e[i].v,x);
17     int t=0,lst=x;for(int i=hd[x];i;i=e[i].nxt)if(e[i].v!=p)v[++t]=i;
18     if(!t)return;g[++cc]=(G){e[v[1]].v,x,e[v[1]].a,e[v[1]].b};
19     for(int i=2;i<t;i++)g[++cc]=(G){lst,++n,0,0},g[++cc]=(G){e[v[i]].v,lst=n,e[v[i]].a,e[v[i]].b};
20     if(t>1)g[++cc]=(G){e[v[t]].v,lst,e[v[t]].a,e[v[t]].b};
21 }
22 void gtrt(int x,int p){sz[x]=1;for(int i=hd[x],y,t;i;i=e[i].nxt)if((y=e[i].v)!=p&&!vis[i]){gtrt(y,x);sz[x]+=sz[y];if((t=max(sz[y],S-sz[y]))<res)res=t,rt=i;}}
23 void gt(int x,int p,int t,st*s,int&tp,ll s1,ll s2){if(t)s[++tp]=(st){s1,s2};for(int i=hd[x],y;i;i=e[i].nxt)if((y=e[i].v)!=p&&!vis[i])gt(y,x,e[i].a+e[i].b,s,tp,s1+e[i].a,s2+e[i].b);}
24 inline void build(st*s,int&tp,double*S,int f)
25 {
26     int t=0;sort(s+1,s+tp+1);
27     for(int i=1;i<=tp;i++)if(i==1||s[i].k!=s[i-1].k){while(t>1&&chk(s[t-1],s[t],s[i]))t--;s[++t]=s[i];}
28     tp=t;if(f)for(int i=1;i<=tp;i++)if(i<tp)S[i]=1.0*(s[i].b-s[i+1].b)/(s[i+1].k-s[i].k);
29 }
30 void sol(int x)
31 {
32     if(sz[x]==1)return;S=sz[x];res=1<<30;gtrt(x,0);
33     int A=e[rt].v,B=e[rt^1].v;vis[rt]=vis[rt^1]=1;s1=s2=0;
34     if(sz[A]>sz[B])sz[A]=S-sz[B];else sz[B]=S-sz[A];
35     gt(A,B,1,p,s1,e[rt].a,e[rt].b);gt(B,A,1,q,s2,0,0);build(p,s1,P,1);build(q,s2,Q,1);
36     for(int i=1,j=1;i<=s1;i++)
37     {
38         while(true)
39         {
40             r[++cnt]=(st){p[i].k+q[j].k,p[i].b+q[j].b};
41             if(j==s2||(i<s1&&P[i]<Q[j]))break;j++;
42         }
43     }
44     sol(A);sol(B);
45 }
46 int main()
47 {
48     int T,Cs=0;scanf("%d",&T);
49     while(T--)
50     {
51         scanf("%d%d",&n,&m);tt=0;cc=0;cnt=0;
52         for(int i=1,u,v,w;i<n;i++)scanf("%d%d%d",&u,&v,&w),add(u,v,1,w);
53         dfs(1,0);tt=1;for(int i=1;i<=n;i++)hd[i]=0;
54         for(int i=1;i<=cc;i++)add(g[i].u,g[i].v,g[i].a,g[i].b);
55         sz[1]=n;sol(1);build(r,cnt,P,0);
56         for(int i=1;i<=n;i++)hd[i]=0;
57         for(int i=1;i<=m;i++)scanf("%lld",&W[i].w),W[i].i=i;
58         sort(W+1,W+m+1,cmp);
59         for(int i=1,j=1;i<=m;i++)
60         {
61             while(j<cnt&&r[j].k*W[i].w+r[j].b<r[j+1].k*W[i].w+r[j+1].b)j++;
62             ans[W[i].i]=r[j].k*W[i].w+r[j].b;
63         }
64         printf("Case #%d:
",++Cs);
65         for(int i=1;i<=m;i++)printf("%lld
",ans[i]);
66     }
67     return 0;
68 }

View Code

gym101955EThe Kouga Ninja Scrolls

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=100005;
 5 const ll inf=1ll<<50;
 6 int n,m,ac[N];ll ax[N],ay[N];
 7 struct st
 8 {
 9     ll f,s,c;
10     inline ll operator+(const st&b){return c==b.c?max(f+b.s,s+b.f):f+b.f;}
11     inline st operator*(const st&b){return (c==b.c)?(st){max(f,b.f),max(s,b.s),c}:(f<b.f?(st){b.f,max(f,b.s),b.c}:(st){f,max(s,b.f),c});} 
12     inline void stv(ll x,int w){f=x;s=-inf;c=w;}
13 };
14 struct D
15 {
16     st a[4];
17     inline D operator+(const D&b){D r;for(int i=0;i<4;i++)r.a[i]=a[i]*b.a[i];return r;}
18     inline void stv(int l){for(int i=0;i<4;i++)a[i].stv(ax[l]*(i&1?1:-1)+ay[l]*(i&2?1:-1),ac[l]);}
19     inline ll calc(){ll r=0;for(int i=0;i<4;i++)r=max(r,a[i]+a[3^i]);return r;}
20 };
21 struct tr{int l,r,m;D s;}T[N<<2];
22 inline void up(int x){T[x].s=T[x<<1].s+T[x<<1|1].s;}
23 void build(int x,int l,int r)
24 {
25     T[x].l=l;T[x].r=r;T[x].m=l+r>>1;if(l==r){T[x].s.stv(l);return;}
26     build(x<<1,l,T[x].m);build(x<<1|1,T[x].m+1,r);up(x);
27 }
28 void upd(int x,int p){if(T[x].l==T[x].r){T[x].s.stv(T[x].l);return;}p<=T[x].m?upd(x<<1,p):upd(x<<1|1,p);up(x);}
29 D qry(int x,int l,int r)
30 {
31     if(l<=T[x].l&&T[x].r<=r)return T[x].s;
32     if(l>T[x].m)return qry(x<<1|1,l,r);
33     if(r<=T[x].m)return qry(x<<1,l,r);
34     return qry(x<<1,l,r)+qry(x<<1|1,l,r);
35 }
36 int main()
37 {
38     int T,Cs=0;scanf("%d",&T);
39     while(T--)
40     {
41         printf("Case #%d:
",++Cs);scanf("%d%d",&n,&m);
42         for(int i=1;i<=n;i++)scanf("%lld%lld%d",&ax[i],&ay[i],&ac[i]);
43         build(1,1,n);
44         while(m--)
45         {
46             int o,k,x,y;scanf("%d%d%d",&o,&k,&x);
47             if(o==1)scanf("%d",&y),ax[k]+=x,ay[k]+=y;
48             if(o==2)ac[k]=x;
49             if(o==3)printf("%lld
",max(0ll,qry(1,k,x).calc()));
50             else upd(1,k);
51         }
52     }
53     return 0;
54 }

View Code

gym101955FCounting Sheep in Ami Dongsuo

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=10005,M=1205,P=1e9+7; 
 4 int inv[M],ifac[M],n,m,w,to[N*3],nxt[N*3],hd[N],d[N],Q[N],pw[M][M],f[N][3][M],tot[M],tmp[M],ans[M];
 5 int main()
 6 {
 7     inv[1]=ifac[0]=1;
 8     for(int i=2;i<M;i++)inv[i]=1ll*inv[P%i]*(P-P/i)%P;
 9     for(int i=1;i<M;i++)ifac[i]=1ll*ifac[i-1]*inv[i]%P;
10     for(int i=1;i<M;i++)for(int j=pw[i][0]=1;j<M;j++)pw[i][j]=1ll*pw[i][j-1]*i%P;
11     int T,Cs=0;scanf("%d",&T);
12     while(T--)
13     {
14         scanf("%d%d%d",&n,&m,&w);w*=3;
15         for(int i=1;i<=n;i++)
16         {
17             int v;scanf("%d",&v);hd[i]=d[i]=0;
18             for(int j=0;j<3;j++)for(int k=1;k<=w;k++)f[i][j][k]=pw[k][v*(j+1)];
19         }
20         for(int i=1,u,v;i<=m;i++){scanf("%d%d",&u,&v);to[i]=v;nxt[i]=hd[u];hd[u]=i;d[v]++;}
21         int L=0,R=0;for(int i=1;i<=n;i++)if(!d[i])Q[++R]=i;
22         while(L<R)for(int x=Q[++L],i=hd[x];i;i=nxt[i])if(!--d[to[i]])Q[++R]=to[i];
23         memset(tot,0,sizeof(tot));
24         for(int i=n;i;i--)
25         {
26             int x=Q[i];
27             for(int t=hd[x];t;t=nxt[t])
28             {
29                 int y=to[t];
30                 for(int j=0;j<3;j++)for(int k=0;k<=w;k++)f[x][j][k]=(f[x][j][k]+f[y][j][k])%P;
31             }
32             for(int j=0;j<=w;j++)
33             {
34                 int res=1ll*f[x][0][j]*f[x][0][j]%P*f[x][0][j]%P;
35                 res=(res-3ll*f[x][0][j]*f[x][1][j]%P+P)%P;
36                 res=(res+2ll*f[x][2][j])%P;tot[j]=(tot[j]+res)%P;
37             }
38         }
39         for(int i=1;i<=w;i++)tot[i]=1ll*tot[i]*inv[6]%P;
40         memset(tmp,0,sizeof(tmp));memset(ans,0,sizeof(ans));tmp[1]=1;
41         for(int i=1;i<=w;i++)for(int j=i+1;j;j--)tmp[j]=(1ll*tmp[j]*(P-i)+tmp[j-1])%P;
42         for(int i=1;i<=w;i++)
43         {
44             for(int j=1;j<=w;j++)tmp[j]=1ll*(tmp[j]-tmp[j-1]+P)*(P-inv[i])%P;
45             int c=1ll*tot[i]*ifac[i]%P*ifac[w-i]%P*(w-i&1?P-1:1)%P;
46             for(int j=1;j<=w;j++)ans[j]=(ans[j]+1ll*tmp[j]*c)%P;
47             for(int j=w;j;j--)tmp[j]=(1ll*tmp[j]*(P-i)+tmp[j-1])%P;
48         }
49         printf("Case #%d:",++Cs);
50         for(int i=1;i<=w;i++)printf(" %d",ans[i]);puts("");
51     }
52     return 0;
53 }

View Code

gym101955GBest ACMer Solves the Hardest Problem

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=13000005,M=6005;
 4 int n,m,px[200005],py[200005],w[M][M];
 5 vector<pair<int,int> >g[N];
 6 inline void chk(int x,int y,int v){if(x>=1&&x<=6000&&y>=1&&y<=6000&&w[x][y])w[x][y]+=v;}
 7 inline void qry(int x,int y,ll&v){if(x>=1&&x<=6000&&y>=1&&y<=6000&&w[x][y])v+=w[x][y];}
 8 int main()
 9 {
10     for(int i=0;i<=6000;i++)for(int j=0;j<=6000;j++)if(i*i+j*j<=10000000)g[i*i+j*j].push_back(make_pair(i,j));else break;
11     int T,Cs=0;scanf("%d",&T);
12     while(T--)
13     {
14         printf("Case #%d:
",++Cs);
15         scanf("%d%d",&n,&m);ll ans=0;
16         for(int i=1;i<=n;i++)scanf("%d%d",&px[i],&py[i]),scanf("%d",&w[px[i]][py[i]]);
17         while(m--)
18         {
19             int o,x,y;scanf("%d%d%d",&o,&x,&y);
20             x=(x+ans)%6000+1;y=(y+ans)%6000+1;
21             if(o==1){scanf("%d",&w[x][y]);n++;px[n]=x;py[n]=y;}
22             else if(o==2)w[x][y]=0;
23             else if(o==3)
24             {
25                 int d,v;scanf("%d%d",&d,&v);
26                 for(int i=0;i<g[d].size();i++)
27                 {
28                     int dx=g[d][i].first,dy=g[d][i].second;
29                     if(!dx&&!dy)chk(x,y,v);
30                     else if(!dx)chk(x,y+dy,v),chk(x,y-dy,v);
31                     else if(!dy)chk(x+dx,y,v),chk(x-dx,y,v);
32                     else chk(x+dx,y+dy,v),chk(x+dx,y-dy,v),chk(x-dx,y+dy,v),chk(x-dx,y-dy,v);
33                 }
34             }
35             else
36             {
37                 int d;scanf("%d",&d);ans=0;
38                 for(int i=0;i<g[d].size();i++)
39                 {
40                     int dx=g[d][i].first,dy=g[d][i].second;
41                     if(!dx&&!dy)qry(x,y,ans);
42                     else if(!dx)qry(x,y+dy,ans),qry(x,y-dy,ans);
43                     else if(!dy)qry(x+dx,y,ans),qry(x-dx,y,ans);
44                     else qry(x+dx,y+dy,ans),qry(x+dx,y-dy,ans),qry(x-dx,y+dy,ans),qry(x-dx,y-dy,ans);
45                 }
46                 printf("%lld
",ans); 
47             }
48         }
49         for(int i=1;i<=n;i++)w[px[i]][py[i]]=0;
50     }
51     return 0;
52 }

View Code

gym101955HRainbow Graph

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gym101955IDistance Between Sweethearts

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=2049;
 5 int mx,sz,p=0,q=1,A[3],B[3],f[3][N][N];ll g[2][N];
 6 inline void fwt(int n,int*a,int f){int w=f==1?0:1;for(int t=1;t<n;t<<=1)for(int i=0;i<n;i+=(t<<1))for(int j=0;j<t;j++){int u=a[i+j],v=a[i+j+t];a[i+j]=(u+v)>>w;a[i+j+t]=(u-v)>>w;}}
 7 inline void fwt(int n,ll*a,int f){int w=f==1?0:1;for(int t=1;t<n;t<<=1)for(int i=0;i<n;i+=(t<<1))for(int j=0;j<t;j++){ll u=a[i+j],v=a[i+j+t];a[i+j]=(u+v)>>w;a[i+j+t]=(u-v)>>w;}}
 8 int main()
 9 {
10     int T,Cs=0;scanf("%d",&T);
11     while(T--)
12     {
13         mx=0;sz=1;memset(f,0,sizeof(f));memset(g[p],0,sizeof(g[0]));
14         for(int i=0;i<3;i++)scanf("%d",&A[i]);
15         for(int i=0;i<3;i++)scanf("%d",&B[i]);
16         for(int i=0;i<3;i++)
17         {
18             mx=max(mx,max(A[i],B[i]));for(;sz<=mx;sz<<=1);
19             for(int x=0;x<=A[i];x++)for(int y=0;y<=B[i];y++)f[i][abs(x-y)][x^y]++;
20         } 
21         ll ans=0;
22         for(int i=0;i<=mx;i++)
23         {
24             for(int j=0;j<3;j++)
25             {
26                 fwt(sz,f[j][i],1);
27                 if(i)for(int k=0;k<sz;k++)f[j][i][k]+=f[j][i-1][k];
28             }
29             swap(p,q);
30             for(int j=0;j<sz;j++)g[p][j]=1ll*f[0][i][j]*f[1][i][j]*f[2][i][j];
31             fwt(sz,g[p],-1);ll res=0;
32             for(int j=sz-1;j>0;j--){int k=i^j;res+=g[p][k]-g[q][k];ans+=res;}
33         }
34         printf("Case #%d: %llu
",++Cs,ans);
35     }
36     return 0;
37 }

View Code

gym101955JHow Much Memory Your Code Is Using?

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int T,Cs=0;scanf("%d",&T);
 6     while(T--)
 7     {
 8         int n,ans=0,c;cin>>n;string s;getline(cin,s);
 9         for(int i=0;i<n;i++)
10         {
11             string s;int t;getline(cin,s);string str=s;
12             if(s[0]=='b'||s[0]=='c')t=1;
13             else if(s[0]=='i'||s[0]=='f')t=4;
14             else if(s[0]=='d')t=8;else if(s[0]=='_')t=16;
15             else if(s[0]=='l'){if(s[5]=='l')t=8;else if(s[5]=='d')t=16;}
16             int p1=str.find("[");c=0;
17             if(p1!=-1){int p2=str.find("]");for(int i=p1+1;i<p2;i++)c=c*10+s[i]-'0';}
18             if(c)t*=c;ans+=t;
19         }
20         ans=ans/1024+(ans%1024>0);
21         cout<<"Case #"<<++Cs<<": "<<ans<<endl;
22     }
23     return 0;
24 }

View Code

gym101955KLet the Flames Begin

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 int main()
 5 {
 6     int T,Cs=0;ll n,m,k;scanf("%d",&T);
 7     while(T--)
 8     {
 9         scanf("%lld%lld%lld",&n,&m,&k);
10         if(k==1)printf("Case #%d: %lld
",++Cs,m);
11         else
12         {
13             ll c=n-m+1,res=(k-1)%c;
14             while(c<n){ll d=min(max((c-res)/(k-1),1ll),n-c);c+=d;res=(res+d*K)%c;} 
15             printf("Case #%d: %lld
",++Cs,res+1);
16         }
17     }
18     return 0;
19 }

View Code

gym101955LMachining Disc Rotors

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=205;
 4 const double eps=1e-9,pi=acos(-1.0);
 5 double ans,X[N],Y[N],L[N],R[N],a[N];
 6 inline int sgn(double x){return x>eps?1:(x<-eps?-1:0);}
 7 int main()
 8 {
 9     int T,Cs=0,n,cc,Ra;scanf("%d",&T);
10     while(T--)
11     {
12         scanf("%d%d",&n,&Ra);cc=0;ans=0;
13         for(int i=1;i<=n;i++)
14         {
15             int x,y,r;scanf("%d%d%d",&x,&y,&r);int t=x*x+y*y;
16             if(t>=(r+Ra)*(r+Ra)||(Ra>r&&t<=(Ra-r)*(Ra-r)))continue;
17             double a1=atan2(y,x),a2=acos((Ra*Ra+t-r*r)/(2*Ra*sqrt(t)));
18             X[++cc]=Ra*cos(a1-a2);Y[cc]=Ra*sin(a1-a2);L[cc]=a1-a2;R[cc]=a1+a2;a[cc]=a1-a2;
19             X[++cc]=Ra*cos(a1+a2);Y[cc]=Ra*sin(a1+a2);a[cc]=a1+a2;
20         }
21         for(int i=1,j;i<=cc;i++)
22         {
23             for(j=1;j<=cc;j+=2)if((sgn(L[j]-a[i]-pi)<0&&sgn(R[j]-a[i]-pi)>0)||(sgn(L[j]-a[i]+pi)<0&&sgn(R[j]-a[i]+pi)>0))break;
24             if(j>cc)ans=Ra*2;
25         }
26         for(int i=1;i<=cc;i++)for(int j=i+1;j<=cc;j++)ans=max(ans,sqrt((X[i]-X[j])*(X[i]-X[j])+(Y[i]-Y[j])*(Y[i]-Y[j])));
27         printf("Case #%d: %.10f
",++Cs,ans);
28     }
29     return 0;
30 }

View Code

gym101955MRenaissance Past in Nancy

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=10005,M=1005,P=1e9+7;
 4 int n,m,ans,f[N][M],g[N][M];
 5 inline void ins(int*f,int x){for(int i=M-1;i>=x;i--)f[i]=(f[i]-f[i-x]+P)%P;}
 6 inline void del(int*f,int x){for(int i=x;i<M;i++)f[i]=(f[i]+f[i-x])%P;}
 7 int main()
 8 {
 9     int T,Cs=0;scanf("%d",&T);
10     while(T--)
11     {
12         printf("Case #%d:
",++Cs);scanf("%d%d",&n,&m);ans=0;
13         memset(f[0],0,sizeof(f[0]));f[0][0]=1;
14         memset(g[0],0,sizeof(g[0]));g[0][0]=1;
15         for(int i=1;i<=n;i++)
16         {
17             int a,b;scanf("%d%d",&a,&b);a++;
18             memcpy(f[i],f[i-1],sizeof(f[i]));
19             memcpy(g[i],g[i-1],sizeof(g[i]));
20             del(f[i],b);ins(g[i],b);
21             if(a*b<M)ins(f[i],a*b),del(g[i],a*b);
22         }
23         for(int i=1;i<=n;i++)for(int j=1;j<M;j++)f[i][j]=(f[i][j]+f[i][j-1])%P;
24         while(m--)
25         {
26             int l,r,c;scanf("%d%d%d",&l,&r,&c);l=(l+ans)%n+1;r=(r+ans)%n+1;if(l>r)swap(l,r);l--;ans=0;
27             for(int i=0;i<=c;i++)ans=(ans+1ll*f[r][i]*g[l][c-i])%P;printf("%d
",ans);
28         }
29     }
30     return 0;
31 }

View Code

loj#507. 「LibreOJ NOI Round #1」接竹竿

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=1000005;
 5 ll n,k,a[N],s[N],f[N],g[N];
 6 int main()
 7 {
 8     scanf("%lld%lld",&n,&k);
 9     for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
10     for(int i=1;i<=n;i++)scanf("%lld",&s[i]),s[i]+=s[i-1];
11     for(int i=n;i;i--)f[i]=max(f[i+1],g[a[i]]-s[i-1]),g[a[i]]=max(g[a[i]],s[i]+f[i+1]);
12     printf("%lld
",f[1]);
13     return 0;
14 }

View Code

loj#508. 「LibreOJ NOI Round #1」失控的未来交通工具

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=1000005;
 5 int n,m,q,f[N],d[N],g[N];
 6 inline int rd()
 7 {
 8     int x=0;char c=getchar();while(!isdigit(c))c=getchar();
 9     while(isdigit(c))x=x*10+c-48,c=getchar();return x;
10 }
11 inline int fnd(int x){if(f[x]==x)return x;fnd(f[x]);d[x]=(d[x]+d[f[x]])%m;return f[x]=f[f[x]];}
12 inline void exgcd(int a,int b,int&x,int&y){if(!b){x=1;y=0;}else {exgcd(b,a%b,y,x);y-=a/b*x;}}
13 inline int inv(int a,int b){int x,y;exgcd(a,b,x,y);return (x%b+b)%b;}
14 int main()
15 {
16     n=rd();m=rd();q=rd();
17     for(int i=1;i<=n;i++)f[i]=i,g[i]=m;
18     while(q--)
19     {
20         int o,u,v,x,y,z,w,t;o=rd();u=rd();v=rd();x=rd();
21         if(o==1)
22         {
23             if((y=fnd(u))^(z=fnd(v)))d[z]=((d[u]+d[v])%m+x)%m,g[y]=__gcd(__gcd(g[y],g[z]),x*2),f[z]=y;
24             else g[y]=__gcd(g[y],((d[u]+d[v])%m+x)%m);
25         }
26         else
27         {
28             y=rd();z=rd();if(fnd(u)!=fnd(v)){puts("0");continue;}
29             w=g[fnd(u)];t=__gcd(w,y);x=((d[u]+d[v])%m+x)%m;if(x%t){puts("0");continue;}
30             x/=t;y/=t;w/=t;printf("%d
",(int)((z-1ll*x*(w-inv(y%w,w))%w+w-1)/w));
31         }
32     }
33     return 0;
34 }

View Code

loj#509. 「LibreOJ NOI Round #1」动态几何问题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int N=130000000;
 5 int cc,pr[10000005],g[N];char mu[N];short f[N];
 6 bitset<N>vis;
 7 inline ll sqr(ll n){return sqrt(n+0.5);}
 8 inline ll qry(ll n)
 9 {
10     if(n<N&&g[n])return g[n];
11     ll i=1,s=0;for(;i*i*i<=n;i++)s+=n/(i*i)*mu[i];
12     while(i*i<=n){ll j=sqr(n/(n/(i*i)));s+=n/(i*i)*(f[j]-f[i-1]);i=j+1;}return s;
13 }
14 inline void init(int n)
15 {
16     mu[1]=f[1]=g[1]=1;
17     for(int i=2;i<=n;i++)
18     {
19         if(!vis[i]){pr[cc++]=i;f[i]=-1;}
20         for(int j=0;j<cc;j++)
21         {
22             int t=pr[j];if(i*t>n)break;
23             vis[i*t]=1;if(i%t==0)break;
24             f[i*t]=-f[i];
25         }
26         mu[i]=f[i];g[i]=g[i-1]+!!f[i];f[i]+=f[i-1];
27     }
28 }
29 int main()
30 {
31     ll n,m,ans=0;scanf("%lld%lld",&n,&m);if(n>m)swap(n,m);init(sqr(n));
32     for(ll i=1,j,s,t;i<=n;i=j+1,s=t){ll x=sqr(n/i),y=sqr(m/i);j=min(n/(x*x),m/(y*y));t=qry(j);ans+=x*y*(t-s);}
33     printf("%lld
",ans);return 0;
34 }

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原文地址：https://www.cnblogs.com/alonefight/p/10927412.html