07 2017 档案
摘要:#include<bits/stdc++.h> using namespace std; int n,a[1010],t,b[1010]; string m; void add(){ for(int i=1;i<=a[0];i++)b[a[0]-i+1]=a[i]; for(int i=1;i<=a
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摘要:#include<bits/stdc++.h>using namespace std;typedef long long LL; void exgcd(int a,int b,LL &x,LL &y){ if(b==0){x=1, y=0; return;} LL tx,ty; exgcd(b,a%
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摘要:#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1e6 + 7;int a[N];LL ans,all; void dfs(int x, LL y){ if(x>a[0]){ans+=all/
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摘要:#include<bits/stdc++.h>using namespace std;const int N=2;const int MOD=10000;struct MAT{int a[N][N];}; MAT operator*(MAT x, MAT y){ MAT z; for(int i=0
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摘要:#include<bits/stdc++.h>using namespace std;#define N 100010#define M 2000010int root[N],R[M],L[M],s[M],m,x,y,n,k,a[N],cnt; void update(int &A,int &B,i
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摘要:#include<bits/stdc++.h> using namespace std; struct node{char pa,data,lc,rc;}a[10001]; int vis[201],root=-1; int xianxu(int i){ printf("%c",i); if(a[i
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摘要:#include<bits/stdc++.h> using namespace std; int n,m,a[110][110],maxn; int min(int a,int b,int c){return min(a,min(b,c));} int main(){ scanf("%d%d",&n
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摘要:#include<bits/stdc++.h> using namespace std; struct point{double x,y;}a[20]; int n; double all=1<<30 ; bool use[20]; void ( double x,double y,int sum,
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摘要:#include<bits/stdc++.h>using namespace std;#define N 500010*2struct edge{int v,next;}e[N];struct qedge{int u,v,next,lca,num;}e1[N];int n,m,t,head[N],t
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摘要:#include<bits/stdc++.h> using namespace std; int f[1000010],n,m; char p[1000010],t[1000010]; void getfail(){ f[0]=f[1]=0; for(int i=1;i<m;i++){ int j=
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