F t 组数据 n 然后n个数
恰好要去掉一个数 使得序列递增 或者递减 （不严格）
画2个图模拟 或者nlong(n)求最长上升子序列
#include<stdio.h> #include<algorithm> #include<string.h> #include<iostream> #include<math.h> #include<queue> #include<stdlib.h> #include<vector> #include<map> using namespace std; #define MAXN 100100 #define inf 1000000007 #define ll long long int z[MAXN]; int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&z[i]); int ok=0; int cnt=0; z[0]=inf; z[n+1]=-1; int i; for(i=2;i<=n;i++) { if(z[i]>z[i-1])//递减 { if(ok==1) { ok=0; break; } if(z[i-1]>=z[i+1]) ok=1; else { if(z[i-2]>=z[i]) ok=1; else { ok=0; break; } } } } if(ok==0&&i==n+1) ok=1; if(ok==1) { printf("YES "); continue; } z[0]=-1; z[n+1]=inf; for(i=2;i<=n;i++) { if(z[i]<z[i-1])//递减 { if(ok==1) { ok=0; break; } if(z[i-1]<=z[i+1]) ok=1; else { if(z[i-2]<=z[i]) ok=1; else { ok=0; break; } } } } if(ok==0&&i==n+1) ok=1; if(ok==1) printf("YES "); else printf("NO "); } return 0; }
L t组数据 n m 然后m*n矩阵 求表面积 最下面不算
找相邻2个柱子之间的关系
#include<stdio.h> #include<algorithm> #include<string.h> #include<iostream> #include<math.h> #include<queue> #include<stdlib.h> #include<vector> #include<map> using namespace std; #define MAXN 110 #define inf 1000000007 #define ll long long int z[MAXN][MAXN]; int main() { int t; scanf("%d",&t); while(t--) { int n,m; memset(z,0,sizeof(z)); scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&z[i][j]); int ans=0; for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { if(z[i][j-1]<z[i][j]) { ans=ans+z[i][j]-z[i][j-1]; } if(z[i][j+1]<z[i][j]) { ans=ans+z[i][j]-z[i][j+1]; } } } for(int i=1;i<=m;i++) { for(int j=1;j<=n;j++) { if(z[j][i]>z[j-1][i]) ans=ans+z[j][i]-z[j-1][i]; if(z[j+1][i]<z[j][i]) ans=ans+z[j][i]-z[j+1][i]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) if(z[i][j]) ans++; } printf("%d ",ans); } return 0; }
J
n^3
#include <iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<math.h> using namespace std ; #define ll __int64 #define MAXN 1010 #define inf 1000000007 #define exp 1e-8 int z[MAXN]; int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&z[i]); int ans=0; for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int a=z[i]+z[j]; for(int k=1;k<=n;k++) { if(i==k||j==k) continue; int b=a^z[k]; if(b>ans) ans=b; } } } printf("%d ",ans); } return 0; }
G 求是不是正多边形 极角排序 相邻的2点构成的边 边长相等 然后相邻的边形成的夹角相等
#include <iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<math.h> using namespace std ; #define ll __int64 #define MAXN 310 #define inf 1000000007 #define exp 1e-8 struct point { public: int x,y; int operator ^ (point b) { return x*b.y-y*b.x; } point operator -(point b) { point c; c.x=x-b.x; c.y=y-b.y; return c; } }res[MAXN]; double dis(point a,point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } bool cmp(point a,point b) { int tmp=( ((a-res[0]) ^ (b-res[0]))); if(tmp>0) return 1; if(tmp==0&&dis(a,res[0])<dis(b,res[0])) return 1; return 0; } int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d%d",&res[i].x,&res[i].y); int c1=0; for(int i=1;i<n;i++) { if(res[i].x<res[c1].x) c1=i; if(res[i].x==res[c1].x&&res[i].y<res[c1].y) c1=i; } swap(res[0],res[c1]); sort(res+1,res+n,cmp); res[n]=res[0]; res[n+1]=res[1]; double d1=dis(res[0],res[1]); point x,y; x=res[1]-res[0]; y=res[2]-res[1]; double dd=x.x*y.x+x.y*y.y; double c2=dd/(sqrt(x.x*x.x+x.y*x.y)*sqrt(y.x*y.x+y.y*y.y)); int ok=0; for(int i=0;i<n;i++) { double d2,d3; d2=dis(res[i],res[i+1]); d3=dis(res[i+1],res[i+2]); if(fabs(d1-d2)>exp||fabs(d1-d3)>exp) ok=1; x=res[i+1]-res[i]; y=res[i+2]-res[i+1]; double dd1=x.x*y.x+x.y*y.y; double c11=dd/(sqrt(x.x*x.x+x.y*x.y)*sqrt(y.x*y.x+y.y*y.y)); if(fabs(c11-c2)>exp) ok=1; } if(ok==0) printf("YES "); else printf("NO "); } return 0; }
K an=2*a(n-2) +a(n-1) + n^4 快速幂
7*7矩阵
#include <iostream> #include<string.h> #include<stdio.h> #include<algorithm> #include<math.h> using namespace std ; #define ll long long #define MAXN 1010 #define inf 2147493647 #define exp 1e-8 struct node { ll z[7][7]; }; node mou(node a,node b,ll c) { node ans; memset(ans.z,0,sizeof(ans.z)); for(int i=0;i<7;i++) { for(int j=0;j<7;j++) { ll sum=0; for(int k=0;k<7;k++) { sum=(sum+a.z[i][k]*b.z[k][j])%c; } ans.z[i][j]=sum; } } return ans; } node quick(node a,ll b,ll c) { node ans; memset(ans.z,0,sizeof(ans.z)); for(int i=0;i<7;i++) ans.z[i][i]=1; while(b>0) { if(b&1) ans=mou(ans,a,c); b>>=1; a=mou(a,a,c); } return ans; } int main() { int t; scanf("%d",&t); while(t--) { ll n,d,b; scanf("%lld%lld%lld",&n,&d,&b); if(n==1) printf("%lld ",d); else if(n==2) printf("%lld ",b); else { node a; memset(a.z,0,sizeof(a.z)); a.z[0][1]=2; a.z[5][4]=2; a.z[1][0]=a.z[1][1]=a.z[2][1]=a.z[2][2]=a.z[3][3]=a.z[4][4]=a.z[5][5]=a.z[6][2]=a.z[6][3]=a.z[6][4]=a.z[6][5]=a.z[6][6]=1; a.z[4][3]=a.z[5][3]=3; a.z[3][2]=a.z[5][2]=4; a.z[4][2]=6; a.z[3][1]=4; a.z[4][1]=6; a.z[5][1]=4; a.z[6][1]=1; node c=quick(a,n-2,inf); int x[7]={0,0,16,8,4,2,1}; x[0]=d; x[1]=b; ll ans=0; for(int i=0;i<7;i++) ans=(ans+(x[i]*c.z[i][1])%inf)%inf; printf("%lld ",ans); } } return 0; }