训练记录
辣鸡lfw和傻逼byf互相表演，都没看出直接暴力跑B得快速幂是1e6*1e6的复杂度，写了一万年，lfw还因为搞错了a,b顺序半天过不了样例，10次方的快速幂byf自己讲课的课件上有，然后别的队看着他课件写得，他自己不会，需要好好复习。导致后面很多很水的题没写。难顶，以后不要3个人同时做一道题。
题解
比赛链接：https://ac.nowcoder.com/acm/contest/885#question
A digits 2
输出n遍n
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int n; 5 6 int main() 7 { 8 int t; 9 scanf("%d",&t); 10 for(int i=1;i<=t;i++) 11 { 12 scanf("%d",&n); 13 for(int j=1;j<=n;j++) 14 printf("%d",n); 15 puts(""); 16 } 17 return 0; 18 }
B generator 1
题解：https://blog.csdn.net/liufengwei1/article/details/98115711
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int size=1e6+5; 5 char n[size]; 6 int x0,x1,a,b; 7 typedef long long LL; 8 LL mod; 9 struct mat 10 { 11 int m[3][3]; 12 mat(){for(int i=0;i<3;i++)for(int j=0;j<3;j++) m[i][j]=0;} 13 friend mat operator*(mat x,mat y) 14 { 15 mat ans; 16 for(int i=1;i<=2;i++) 17 { 18 for(int j=1;j<=2;j++) 19 { 20 for(int k=1;k<=2;k++) 21 { 22 ans.m[i][k] = (ans.m[i][k]+1LL*x.m[i][j]*y.m[j][k])%mod; 23 } 24 } 25 } 26 return ans; 27 } 28 }; 29 inline mat quick_pow(mat a,int b) 30 { 31 mat ans; 32 ans.m[1][1]=1;ans.m[2][2]=1; 33 while(b) 34 { 35 if(b&1) ans=ans*a; 36 a=a*a; 37 b>>=1; 38 } 39 return ans; 40 } 41 int main() 42 { 43 scanf("%d%d%d%d",&x0,&x1,&a,&b); 44 scanf("%s%lld",n,&mod); 45 mat ori; 46 int len =strlen(n); 47 ori.m[1][2]=1,ori.m[2][1]=b;ori.m[2][2]=a; 48 mat ans; 49 ans.m[1][1]=1;ans.m[2][2]=1; 50 for(int i=len-1;i>=0;i--) 51 { 52 ans=ans*quick_pow(ori,n[i]-'0'); 53 ori=quick_pow(ori,10); 54 } 55 printf("%lld ",(1LL*x0*ans.m[1][1]+1LL*x1*ans.m[1][2])%mod); 56 }
C generator 2
https://blog.csdn.net/baiyifeifei/article/details/98263274
#include<bits/stdc++.h> using namespace std; typedef long long LL; int quick_pow(int a,int b,int mod) { int ans=1; while(b) { if(b&1) ans=1LL*ans*a%mod; a=1LL*a*a%mod; b>>=1; } return ans; } LL n; int x0,a,b,p; LL v; #define int long long unordered_map<LL,LL> mp; int loop,up; void pre_BSGS(int p,int b) { mp.clear(); up=ceil(pow(p,2.0/3)); int t=1; for(int i=0;i<=up;i++) { if(i==up) loop=t; mp[t]=i; t=1LL*t*b%p; } } int BSGS(int B,int N,int P) { int m=ceil(pow(p,1.0/3)); int obj=quick_pow(N,P-2,P); for(int i=1;i<=m;i++) { obj=1LL*obj*loop%P; if(mp.count(obj)) { return 1LL*i*up-mp[obj]; } } return -1; } #undef int int main() { int t; int q; // freopen("c.in","r",stdin); scanf("%d",&t); while(t--) { scanf("%lld%d%d%d%d",&n,&x0,&a,&b,&p); LL ab_apo=1LL*b*quick_pow(a-1,p-2,p)%p; LL x0ab_apo=(x0+ab_apo)%p; LL o_b=quick_pow(b,p-2,p); pre_BSGS(p,a); scanf("%d",&q); while(q--) { scanf("%lld",&v); if(a==1) { LL ans=(1LL*(v-x0)*o_b%p+p)%p; printf("%lld ",ans<n?ans:-1); continue; } if(a==0) { if(x0==v) puts("0"); else if(b==v) puts("1"); else puts("-1"); continue; } v=(v+ab_apo)%p; v=1LL*v*quick_pow(x0ab_apo,p-2,p)%p; LL ans=BSGS(a,v,p); printf("%d ",ans<n?ans:-1); } } }
D generator 3
unsolved
E independent set 1
题解：https://blog.csdn.net/liufengwei1/article/details/98207215
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int n,m,sum; 5 int e[27],mi[27]; 6 char dp[1<<26]; 7 8 inline void prework() 9 { 10 scanf("%d%d",&n,&m); 11 int u,v; 12 for(int i=1;i<=m;i++) 13 { 14 scanf("%d%d",&u,&v); 15 e[v]|=(1<<u); 16 e[u]|=(1<<v); 17 } 18 mi[0]=1; 19 for(int i=1;i<=n;i++) 20 mi[i]=mi[i-1]*2; 21 int s=(1<<n)-1; 22 for(int i=0;i<n;i++) 23 e[i]=s^(e[i]|mi[i]); 24 } 25 26 inline void mainwork() 27 { 28 int s=(1<<n)-1; 29 int x,id; 30 for(int i=1;i<=s;i++) 31 { 32 x=i&-i;id=log2(x); 33 dp[i]=max(dp[i^x],(char)(dp[i&e[id]]+1)); 34 sum+=dp[i]; 35 } 36 } 37 38 inline void print() 39 { 40 printf("%d",sum); 41 } 42 43 int main() 44 { 45 prework(); 46 mainwork(); 47 print(); 48 return 0; 49 }
F maximum clique 1
https://blog.csdn.net/baiyifeifei/article/details/98263347
做法1（二分图匹配）
#include<bits/stdc++.h> using namespace std; const int maxn=5e3+5; const int maxm=5e5+5; typedef pair<int,int> pii; int head[maxn],to[maxm]; int nxt[maxm]; int a[maxn]; inline int lowbit(int x){return x&(-x);} int n; bool vis[maxn]; bool color[maxn]; int tot; int un,vn; void init() { tot=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v) { to[tot]=v;nxt[tot]=head[u]; head[u]=tot++; } int linker[maxn]; bool used[maxn]; bool dfs(int u) { for(int i=head[u];i!=-1;i=nxt[i]) { int v=to[i]; if(!used[v]) { used[v]=true; if(linker[v]==-1||dfs(linker[v])) { linker[v]=u; return true; } } } return false; } int hungary() { int res=0; memset(linker,-1,sizeof(linker)); for(int u=1;u<=un;u++) { if(color[u]!=0) continue; memset(used,0,sizeof(used)); if(dfs(u)) res++; } return res; } void colormaker(int u,int col) { vis[u]=true; color[u]=col; for(int i=head[u];i!=-1;i=nxt[i]) { if(vis[to[i]]) continue; colormaker(to[i],~col); } } vector<int> ans; queue<pii> q; void planfinder() { ans.clear(); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) { if(!color[i]) continue; if(linker[i]!=-1) continue; if(vis[i]) continue; while(!q.empty()) q.pop(); q.push(pii(i,0)); while(!q.empty()) { int v=q.front().first; int sw=q.front().second; vis[v]=true; q.pop(); for(int j=head[v];j!=-1;j=nxt[j]) { if(vis[to[j]]) continue; if(sw==0) {if(linker[to[j]]!=v) q.push(pii(to[j],1));} else {if(linker[to[j]]==v||linker[v]==to[j]) q.push(pii(to[j],0));} } } } memset(used,0,sizeof(used)); for(int i=1;i<=n;i++) if((vis[i]&&(!color[i]))||(!vis[i]&&color[i])) used[i]=true; } int main() { init(); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int sign=a[i]^a[j]; if(sign==lowbit(sign)) { addedge(i,j); addedge(j,i); } } } memset(color,0,sizeof(color)); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) { if(!vis[i])colormaker(i,0); } un=vn=n; printf("%d ",n-hungary()); planfinder(); for(int i=1;i<=n;i++) { if(!used[i]) printf("%d ",a[i]); } }
做法2（网络流）
#include<bits/stdc++.h> using namespace std; const int maxn=5005; const int maxm=4e6+5; const int inf=0x3f3f3f3f; struct Edge{ int to,nxt,cap,flow; }edge[maxm]; int tol; int head[maxn]; void init(){ tol=2; memset(head,-1,sizeof(head)); } inline int lowbit(int x){return x&(-x);} void AddEdge(int u,int v,int w,int rw=0){ edge[tol].to=v;edge[tol].cap=w;edge[tol].flow=0; edge[tol].nxt=head[u];head[u]=tol++; edge[tol].to=u;edge[tol].cap=rw;edge[tol].flow=0; edge[tol].nxt=head[v];head[v]=tol++; } int Q[maxn]; int dep[maxn],cur[maxn],sta[maxn]; bool bfs(int s,int t,int n){ int front=0,tail=0; memset(dep,-1,sizeof(dep[0])*(n+1)); dep[s]=0; Q[tail++]=s; while(front<tail){ int u=Q[front++]; for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].to; if(edge[i].cap>edge[i].flow&&dep[v]==-1){ dep[v]=dep[u]+1; if(v==t) return true; Q[tail++]=v; } } } return false; } int dinic(int s,int t,int n){ int maxflow=0; while(bfs(s,t,n)){ for(int i=0;i<n;i++) cur[i]=head[i]; int u=s,tail=0; while(cur[s]!=-1){ if(u==t){ int tp=inf; for(int i=tail-1;i>=0;i--) { tp=min(tp,edge[sta[i]].cap-edge[sta[i]].flow); } maxflow+=tp; for(int i=tail-1;i>=0;i--){ edge[sta[i]].flow+=tp; edge[sta[i]^1].flow-=tp; if(edge[sta[i]].cap-edge[sta[i]].flow==0) tail=i; } u=edge[sta[tail]^1].to; } else if(cur[u]!=-1&&edge[cur[u]].cap>edge[cur[u]].flow&&dep[u]+1==dep[edge[cur[u]].to]){ sta[tail++]=cur[u]; u=edge[cur[u]].to; } else{ while(u!=s&&cur[u]==-1) u=edge[sta[--tail]^1].to; cur[u] = edge [cur[u]].nxt; } } } return maxflow; } bool color[maxn],vis[maxn]; int n; int a[maxn]; void dfs2(int v) { vis[v]=true; for(int i=head[v];i!=-1;i=edge[i].nxt) { if(edge[i].cap>edge[i].flow&&!vis[edge[i].to]) dfs2(edge[i].to); } } void connect(int u,int v) { if(color[u]>color[v]) swap(u,v); AddEdge(u,v,1); } int main() { init(); scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]),color[i]=__builtin_popcount(a[i])%2; for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int sign=a[i]^a[j]; if(sign==lowbit(sign)) { connect(i,j); } } } int s=0,t=n+1; for(int i=1;i<=n;i++) { if(!color[i]) AddEdge(s,i,1); else AddEdge(i,t,1); } printf("%d ",n-dinic(s,t,t+1)); memset(vis,0,sizeof(vis)); dfs2(s); for(int i=1;i<=n;i++) { if(vis[i]) {if(!color[i]) vis[i]=true;else vis[i]=false;} else {if(!color[i]) vis[i]=false;else vis[i]=true;} } for(int i=1;i<=n;i++) { if(vis[i]) printf("%d ",a[i]); } puts(""); }
G subsequence 1
基础DP题，不想写题解= =
1 #include<bits/stdc++.h> 2 #define maxl 3010 3 using namespace std; 4 5 const int mod=998244353; 6 int n,m,ans; 7 int f[maxl][maxl][2]; 8 int dp[maxl][maxl]; 9 char s[maxl],t[maxl]; 10 11 inline void prework() 12 { 13 scanf("%d%d",&n,&m); 14 scanf("%s%s",s+1,t+1); 15 for(int i=0;i<=n;i++) 16 for(int j=0;j<=n;j++) 17 { 18 f[i][j][1]=f[i][j][0]=0; 19 dp[i][j]=0; 20 } 21 for(int i=0;i<=n;i++) 22 f[i][0][0]=1,dp[i][0]=1; 23 } 24 25 inline void mainwork() 26 { 27 for(int i=1;i<=n;i++) 28 { 29 if(s[i]!='0') 30 { 31 if(s[i]>t[1]) 32 { 33 f[i][1][1]=((f[i-1][1-1][1]+f[i-1][1-1][0])%mod+f[i-1][1][1])%mod; 34 f[i][1][0]=f[i-1][1][0]; 35 } 36 else if(s[i]==t[1]) 37 { 38 f[i][1][1]=(f[i-1][1-1][1]+f[i-1][1][1])%mod; 39 f[i][1][0]=(f[i-1][1-1][0]+f[i-1][1][0])%mod; 40 } 41 else 42 f[i][1][1]=f[i-1][1][1],f[i][1][0]=f[i-1][1][0]; 43 } 44 else 45 f[i][1][1]=f[i-1][1][1],f[i][1][0]=f[i-1][1][0]; 46 for(int j=2;j<=m;j++) 47 { 48 if(s[i]>t[j]) 49 { 50 f[i][j][1]=((f[i-1][j-1][0]+f[i-1][j-1][1])%mod+f[i-1][j][1])%mod; 51 f[i][j][0]=f[i-1][j][0]; 52 } 53 if(s[i]==t[j]) 54 { 55 f[i][j][1]=(f[i-1][j-1][1]+f[i-1][j][1])%mod; 56 f[i][j][0]=(f[i-1][j-1][0]+f[i-1][j][0])%mod; 57 } 58 if(s[i]<t[j]) 59 { 60 f[i][j][1]=(f[i-1][j-1][1]+f[i-1][j][1])%mod; 61 f[i][j][0]=f[i-1][j][0]; 62 } 63 } 64 } 65 for(int i=1;i<=n;i++) 66 { 67 if(s[i]!='0') 68 dp[i][1]=(dp[i-1][1]+dp[i-1][1-1])%mod; 69 else 70 dp[i][1]=dp[i-1][1]; 71 for(int j=2;j<=n;j++) 72 dp[i][j]=(dp[i-1][j]+dp[i-1][j-1])%mod; 73 } 74 ans=f[n][m][1]; 75 for(int i=m+1;i<=n;i++) 76 ans=(ans+dp[n][i])%mod; 77 78 } 79 80 inline void print() 81 { 82 printf("%d ",ans); 83 } 84 85 int main() 86 { 87 int t; 88 scanf("%d",&t); 89 for(int i=1;i<=t;i++) 90 { 91 prework(); 92 mainwork(); 93 print(); 94 } 95 return 0; 96 }
H subsequence 2
题解：https://blog.csdn.net/liufengwei1/article/details/98100854
1 #include<bits/stdc++.h> 2 #define maxl 100010 3 using namespace std; 4 5 int n,m; 6 int pos[26][maxl]; 7 int len,mlen[26]; 8 char ch[10]; 9 char s[maxl]; 10 char ans[maxl]; 11 bool in[maxl]; 12 bool flag; 13 14 inline void prework() 15 { 16 scanf("%d%d",&n,&m); 17 int cnt0,cnt1; 18 memset(mlen,-1,sizeof(mlen)); 19 flag=true; 20 for(int i=1;i<=m*(m-1)/2;i++) 21 { 22 scanf("%s%d",ch,&len); 23 cnt0=cnt1=0; 24 if(len>0) 25 { 26 scanf("%s",s+1); 27 for(int j=1;j<=len;j++) 28 { 29 if(s[j]==ch[0]) 30 { 31 cnt0++; 32 pos[ch[0]-'a'][cnt0]+=cnt1; 33 } 34 else 35 { 36 cnt1++; 37 pos[ch[1]-'a'][cnt1]+=cnt0; 38 } 39 } 40 } 41 if(mlen[ch[0]-'a']==-1) 42 mlen[ch[0]-'a']=cnt0; 43 else if(mlen[ch[0]-'a']!=cnt0) 44 flag=false; 45 if(mlen[ch[1]-'a']==-1) 46 mlen[ch[1]-'a']=cnt1; 47 else if(mlen[ch[1]-'a']!=cnt1) 48 flag=false; 49 } 50 } 51 52 inline void mainwork() 53 { 54 if(!flag) return; 55 for(int i=0;i<m;i++) 56 for(int j=1;j<=mlen[i];j++) 57 if(pos[i][j]+j<=n) 58 { 59 if(!in[pos[i][j]+j]) 60 { 61 ans[pos[i][j]+j]='a'+i; 62 in[pos[i][j]+j]=true; 63 } 64 else 65 { 66 flag=false; 67 return; 68 } 69 } 70 else 71 { 72 flag=false; 73 return; 74 } 75 for(int i=1;i<=n;i++) 76 if(!in[i]) 77 { 78 flag=false; 79 return; 80 } 81 } 82 83 inline void print() 84 { 85 if(flag) 86 for(int i=1;i<=n;i++) 87 printf("%c",ans[i]); 88 else 89 puts("-1"); 90 } 91 92 int main() 93 { 94 prework(); 95 mainwork(); 96 print(); 97 return 0; 98 }
I three points 1
题解：https://blog.csdn.net/liufengwei1/article/details/98179088
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const double eps=9e-7; 5 6 double w,h; 7 double a[4]; 8 double b[4]; 9 bool flag; 10 struct node 11 { 12 double x,y; 13 node(double a=0,double b=0) 14 { 15 x=a;y=b; 16 } 17 void transxy(double sinb,double cosb) 18 { 19 double tx=x,ty=y; 20 x=tx*cosb-ty*sinb; 21 y=tx*sinb+ty*cosb; 22 } 23 }ans[4]; 24 25 inline int sgn(double x) 26 { 27 if(x>-eps && x<eps) return 0; 28 if(x>0) return 1; 29 else return -1; 30 } 31 32 inline void prework() 33 { 34 scanf("%lf%lf",&w,&h); 35 for(int i=1;i<=3;i++) 36 scanf("%lf",&a[i]),b[i]=a[i]; 37 } 38 39 inline bool jug(double w,double h,double a,double b,double c) 40 { 41 double sita1=acos((a*a+c*c-b*b)/(2*a*c)); 42 double sita2=max(a>w?acos(w/a):0.0,c>w?acos(w/c)-sita1:0.0); 43 ans[1]=node{a*cos(sita2),a*sin(sita2)}; 44 ans[2]=node{0.0,0.0}; 45 ans[3]=node{c*cos(sita1+sita2),c*sin(sita1+sita2)}; 46 for(int i=1;i<=3;i++) 47 if(ans[i].x<=-eps || ans[i].x>=w+eps || ans[i].y<-eps || ans[i].y>h+eps) 48 return false; 49 return true; 50 } 51 52 inline void mainwork() 53 { 54 int rev; 55 for(rev=0;rev<2;rev++) 56 { 57 if(jug(w,h,a[1],a[2],a[3])) 58 break; 59 if(jug(w,h,a[1],a[3],a[2])) 60 { 61 swap(ans[1],ans[2]); 62 break; 63 } 64 if(jug(w,h,a[2],a[1],a[3])) 65 { 66 swap(ans[2],ans[3]); 67 break; 68 } 69 if(jug(w,h,a[2],a[3],a[1])) 70 { 71 swap(ans[1],ans[2]); 72 swap(ans[2],ans[3]); 73 break; 74 } 75 if(jug(w,h,a[3],a[1],a[2])) 76 { 77 swap(ans[1],ans[3]); 78 swap(ans[2],ans[3]); 79 break; 80 } 81 if(jug(w,h,a[3],a[2],a[1])) 82 { 83 swap(ans[1],ans[3]); 84 break; 85 } 86 swap(w,h); 87 } 88 if(rev) 89 { 90 for(int i=1;i<=3;i++) 91 swap(ans[i].x,ans[i].y); 92 } 93 } 94 95 inline void print() 96 { 97 for(int i=1;i<=2;i++) 98 printf("%.9f %.9f ",ans[i].x,ans[i].y); 99 printf("%.9f %.9f ",ans[3].x,ans[3].y); 100 } 101 102 int main() 103 { 104 int t; 105 scanf("%d",&t); 106 for(int i=1;i<=t;i++) 107 { 108 prework(); 109 mainwork(); 110 print(); 111 } 112 return 0; 113 }
J three points 2
unsolved