Problem 1004
直接用数组模拟
为了避免memset
可以一遍正着统计，一遍倒着统计
注意统计的时候不能出现乘除法
不然会T
然后记录一个cnt
cnt大于2e8就跳出
同时有效数字的位数够了也跳出
当然所有的数都不同以及只有2个数相同要特殊处理
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<algorithm> #include<cstring> #include<string> #include<vector> #include<map> #include<set> #include<queue> using namespace std; const double eps=1e-5; int _,n,a[60010]; int p[60010],q[60010]; int main() { scanf("%d",&_); while (_--) { scanf("%d",&n); int i,j; for (i=1;i<=n;i++) { scanf("%d",&a[i]); p[a[i]]=0,q[a[i]]=0; } int mx=0,cnt=0; for (i=1;i<=n;i++) { if (q[a[i]]) mx=max(mx,i-q[a[i]]); else cnt++; q[a[i]]=i; } if (cnt==1) { printf("%.9f ",1.0/n); continue; } else if (cnt==2) { if (2*(mx-1)<n) printf("%.9f ",2.0/n); else printf("%.9f ",1.0/(mx-1)); continue; } p[a[1]]++; int fz=1; double ans=cnt*1.0/n; cnt=0; for (i=2;i<=n;i++) { int nfz; if (i&1) { if (!p[a[n-i+1]]) fz++; p[a[n-i+1]]++; int pre=n; nfz=fz; for (j=n-i;j>=1;j--) { p[a[pre]]--; if (!p[a[pre]]) fz--; if (!p[a[j]]) fz++; p[a[j]]++; if (fz<nfz) nfz=fz; pre--; cnt++; if (cnt>2e8) break; } } else { if (!p[a[i]]) fz++; p[a[i]]++; int pre=1; nfz=fz; for (j=i+1;j<=n;j++) { p[a[pre]]--; if (!p[a[pre]]) fz--; if (!p[a[j]]) fz++; p[a[j]]++; if (fz<nfz) nfz=fz; pre++; cnt++; if (cnt>2e8) break; } } double as=nfz*1.0/i; //cout<<nfz<<" "<<i<<endl; if (as<ans) ans=as; if (ans<eps) break; if (cnt>2e8) break; } printf("%.9f ",ans); } return 0; }
Problem 1005
取w=min(d1,2,d2,3)w=min(d1,2,d2,3)
因为一条边可以往返跑2ci
所以如果存在距离为d的方案
那么一定存在距离为d+2w的方案
设f[i][j]为从起点出发到i，距离模2w为j时的最短路
根据f[2][j]解不等式即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 #include<functional> 13 using namespace std; 14 typedef long long ll; 15 typedef pair<ll,int> P; 16 const ll inf=1ll<<62; 17 int _,g[4],st; 18 ll f[4][60001],k; 19 priority_queue<P,vector<P>,greater<P> >q; 20 void upd(int x,int y,ll z) 21 { 22 if (z<f[x][y]) 23 { 24 f[x][y]=z; 25 q.push({z,y<<2|x}); 26 } 27 } 28 int main() 29 { 30 scanf("%d",&_); 31 while (_--) 32 { 33 scanf("%lld%d%d%d%d",&k,&g[0],&g[1],&g[2],&g[3]); 34 st=min(g[0],g[1])*2; 35 int i,j; 36 for (i=0;i<4;i++) 37 for (j=0;j<st;j++) f[i][j]=inf; 38 upd(1,0,0); 39 while (!q.empty()) 40 { 41 P t=q.top(); 42 q.pop(); 43 int x=t.second%4; 44 int y=t.second/4; 45 if (f[x][y]<t.first) continue; 46 upd((x+1)%4,(y+g[x])%st,t.first+g[x]); 47 upd((x+3)%4,(y+g[(x+3)%4])%st,t.first+g[(x+3)%4]); 48 } 49 ll ans=inf; 50 for (i=0;i<st;i++) 51 if (f[1][i]<k) 52 { 53 ll sum=f[1][i]+(k-f[1][i])/st*st; 54 while (sum<k) sum+=st; 55 ans=min(ans,sum); 56 } 57 else 58 ans=min(ans,f[1][i]); 59 printf("%lld ",ans); 60 } 61 return 0; 62 }
Problem 1007
度数为1的点匹配方案是唯一的
先拓扑排序去掉所有度数为1的点
剩下的图的每个连痛快是个环
间隔取环上的点
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 typedef long long ll; 14 const ll mod=998244353; 15 int _,n,du[600010],p[600100]; 16 bool vis[600010]; 17 vector<pair<int,int>> a[600010]; 18 int get(int x,int y) 19 { 20 for (auto u:a[x]) 21 if (u.first==y) return u.second; 22 } 23 int main() 24 { 25 scanf("%d",&_); 26 while (_--) 27 { 28 scanf("%d",&n); 29 int i,j; 30 for (i=1;i<=2*n;i++) 31 a[i].clear(); 32 memset(du,0,sizeof(du)); 33 memset(vis,false,sizeof(vis)); 34 for (i=1;i<=n;i++) 35 for (j=0;j<2;j++) 36 { 37 int x,y; 38 scanf("%d%d",&x,&y); 39 a[i].push_back({x+n,y}); 40 a[x+n].push_back({i,y}); 41 du[i]++; 42 du[x+n]++; 43 } 44 n<<=1; 45 queue<int> q; 46 for (i=1;i<=n;i++) 47 if (du[i]==1) 48 q.push(i); 49 ll ans=1; 50 while (!q.empty()) 51 { 52 int x=q.front(); 53 q.pop(); 54 int y; 55 for (auto u:a[x]) 56 if (!vis[u.first]) 57 { 58 y=u.first; 59 ans=1ll*ans*u.second%mod; 60 break; 61 } 62 vis[x]=vis[y]=true; 63 for (auto u:a[y]) 64 if (!vis[u.first]) 65 { 66 du[u.first]--; 67 if (du[u.first]==1) 68 q.push(u.first); 69 } 70 } 71 for (i=1;i<=n;i++) 72 if (!vis[i]) 73 { 74 vis[i]=true; 75 int cnt=1; 76 p[cnt]=i; 77 int cm=i; 78 while (true) 79 { 80 bool fg=true; 81 for (auto u:a[cm]) 82 if (!vis[u.first]) 83 { 84 fg=false; 85 cm=u.first; 86 break; 87 } 88 if (fg) break; 89 //cout<<cm<<" "; 90 vis[cm]=true; 91 p[++cnt]=cm; 92 } 93 //cout<<endl; 94 p[cnt+1]=p[1]; 95 ll x=1,y=1; 96 for (j=1;j<=cnt;j+=2) x=1ll*x*get(p[j],p[j+1])%mod; 97 for (j=2;j<=cnt;j+=2) y=1ll*y*get(p[j],p[j+1])%mod; 98 ans=1ll*ans*(x+y)%mod; 99 } 100 printf("%lld ",ans); 101 } 102 return 0; 103 }
Problem 1011
按照题意模拟即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 const char s[10][8][5]= 14 {{".XX.", 15 "X..X", 16 "X..X", 17 "....", 18 "X..X", 19 "X..X", 20 ".XX."}, 21 22 {"....", 23 "...X", 24 "...X", 25 "....", 26 "...X", 27 "...X", 28 "...."}, 29 30 {".XX.", 31 "...X", 32 "...X", 33 ".XX.", 34 "X...", 35 "X...", 36 ".XX.",}, 37 38 {".XX.", 39 "...X", 40 "...X", 41 ".XX.", 42 "...X", 43 "...X", 44 ".XX." }, 45 46 {"....", 47 "X..X", 48 "X..X", 49 ".XX.", 50 "...X", 51 "...X", 52 "...."}, 53 54 {".XX.", 55 "X...", 56 "X...", 57 ".XX.", 58 "...X", 59 "...X", 60 ".XX.",}, 61 62 {".XX.", 63 "X...", 64 "X...", 65 ".XX.", 66 "X..X", 67 "X..X", 68 ".XX.",}, 69 70 {".XX.", 71 "...X", 72 "...X", 73 "....", 74 "...X", 75 "...X", 76 "...."}, 77 78 { ".XX.", 79 "X..X", 80 "X..X", 81 ".XX.", 82 "X..X", 83 "X..X", 84 ".XX.",}, 85 86 { ".XX.", 87 "X..X", 88 "X..X", 89 ".XX.", 90 "...X", 91 "...X", 92 ".XX.",} 93 }; 94 char c[8][22]; 95 void check(int pos) 96 { 97 int num,i,j; 98 for (num=0;num<10;num++) 99 { 100 bool ok=true; 101 for (i=0;i<7;i++) 102 { 103 for (j=pos;j<=pos+3;j++) 104 if (c[i][j]!=s[num][i][j-pos]) 105 { 106 ok=false; 107 break; 108 } 109 if (!ok) break; 110 } 111 if (ok) 112 { 113 printf("%d",num); 114 return; 115 } 116 } 117 } 118 int main() 119 { 120 int _; 121 scanf("%d",&_); 122 while (_--) 123 { 124 int i,j; 125 for (i=0;i<7;i++) 126 scanf("%s",c[i]); 127 check(0); 128 check(5); 129 printf(":"); 130 check(12); 131 check(17); 132 printf(" "); 133 } 134 return 0; 135 }
Problem 1012
f[i][j][k]表示只考虑a，b两个数组，上升下降状态是k的方案数
直接暴力枚举会T
这时可以建立辅助数组g和h
g[i][y][z]表示从f[x][y][z]开始，要更新的是i的方案数
h[i][j][z]表示从f[x][y][z]开始，已经经过g，要更新的是j的方案数
这样每次就只有一个变量在动
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cstring> 7 #include<string> 8 #include<vector> 9 #include<map> 10 #include<set> 11 #include<queue> 12 using namespace std; 13 const int mod=998244353; 14 int _,n,m,a[2010],b[2010]; 15 int g[2010][2010][2],h[2010][2010][2]; 16 int main() 17 { 18 scanf("%d",&_); 19 while (_--) 20 { 21 scanf("%d%d",&n,&m); 22 int i,j,k; 23 for (i=1;i<=n;i++) scanf("%d",&a[i]); 24 for (i=1;i<=m;i++) scanf("%d",&b[i]); 25 memset(g,0,sizeof(g)); 26 memset(h,0,sizeof(h)); 27 int ans=0; 28 for (i=1;i<=n;i++) 29 for (j=1;j<=m;j++) 30 for (k=0;k<2;k++) 31 { 32 if (a[i]==b[j]) 33 { 34 int t=h[i][j][1-k]; 35 if (!k) t=(t+1)%mod; 36 if (t) 37 { 38 ans=(ans+t)%mod; 39 g[i+1][j][k]=(g[i+1][j][k]+t)%mod; 40 } 41 } 42 if (g[i][j][k]) 43 { 44 g[i+1][j][k]=(g[i+1][j][k]+g[i][j][k])%mod; 45 if (!k) 46 { 47 if (a[i]>b[j]) 48 h[i][j+1][k]=(h[i][j+1][k]+g[i][j][k])%mod; 49 } 50 else 51 { 52 if (a[i]<b[j]) 53 h[i][j+1][k]=(h[i][j+1][k]+g[i][j][k])%mod; 54 55 } 56 } 57 if (h[i][j][k]) h[i][j+1][k]=(h[i][j+1][k]+h[i][j][k])%mod; 58 } 59 printf("%d ",ans); 60 } 61 return 0; 62 }