A. CQXYM Count Permutations

https://codeforces.com/contest/1581/problem/A
就是 3*3*4*...*(2n)

#include <bits/stdc++.h>
using namespace std;

#define Ha 1000000007

long long n;

void solve()
{
	scanf("%lld",&n);
	long long ret=1;
	for (int i=3; i<=n*2; i++)
		ret=ret*i%Ha;
	printf("%lld\n",ret);
}


int main()
{
	int ttt;
	scanf("%d",&ttt);
	while (ttt--) {
		solve();
	}
	return 0;
}

B. Diameter of Graph

https://codeforces.com/contest/1581/problem/B
除了不能成连通图，其它大部分情况都可以（往菊花图那弄就行了），就各种特判咯。

#include <bits/stdc++.h>
using namespace std;



void solve()
{
	long long n,m,k;
	scanf("%lld%lld%lld",&n,&m,&k);
	
	if (n==1) {
		if (m==0 && k>=2)
			puts("YES");
		else
			puts("NO");
		return;
	}
	
	if (n==2) {
		if (m==0)
			puts("NO");
		else if (m==1 && k>=3)
			puts("YES");
		else
			puts("NO");
		return;
	}
	
	if (m<n-1) {
		puts("NO");
		return;
	}
	
	if (m>n*(n-1)/2) {
		puts("NO");
		return;
	}

	if (k>=4) {
		puts("YES");
		return;
	}
	
	if (k==3 && m==n*(n-1)/2) {
		puts("YES");
		return;
	}
	
	puts("NO");
	
	return;
}


int main()
{
	int ttt;
	scanf("%d",&ttt);
	while (ttt--) {
		solve();
	}
	return 0;
}

C - Portal

https://codeforces.com/contest/1581/problem/C
用的是三次方的算法。先枚举矩形上下边，然后右边往右扫，随着右边往右，左边也是往右的（即最优的左边不会在之前的左边）。

#include <bits/stdc++.h>
using namespace std;

#define MAXN 500

int n,m;
char a[MAXN][MAXN];
int g[MAXN][MAXN], sg[MAXN][MAXN];

//g: cnt('1')

int G(int x1, int y1, int x2, int y2)
{
	return sg[x2][y2]-sg[x1-1][y2]-sg[x2][y1-1]+sg[x1-1][y1-1];
}

int fun(int x1, int y1, int x2, int y2)
{
	int ret=0;
	ret+=2*G(x1+1, y1+1, x2-1, y2-1);
	ret-=G(x1, y1, x2, y2);
	ret+=2*(x2-x1+y2-y1);
	ret-=(a[x1][y1]=='0');
	ret-=(a[x1][y2]=='0');
	ret-=(a[x2][y1]=='0');
	ret-=(a[x2][y2]=='0');
	return ret;
}


void solve()
{
	scanf("%d%d",&n,&m);
	for (int i=1; i<=n; i++)
		scanf("%s",a[i]+1);
	
	
	for (int i=1; i<=n; i++)
		for (int j=1; j<=m; j++) {
			g[i][j]=sg[i][j]=0;
		}
	
	for (int i=1; i<=n; i++)
		for (int j=1; j<=m; j++)
			g[i][j]=g[i][j-1]+(a[i][j]=='1');
	
	for (int i=1; i<=n; i++)
		for (int j=1; j<=m; j++)
			sg[i][j]= sg[i-1][j]+g[i][j];
	
	int ans=fun(1,1,5,4), tmp;
	
	
	
	for (int i1=1; i1+4<=n; i1++) {
		for (int i2=i1+4; i2<=n; i2++) {
			int cur=ans;
			int j1=1, j2;
			for (j2=4; j2<=m; j2++) {
				if ((tmp = fun(i1, j1, i2, j2)) < cur) {
					cur=tmp;
				}
				while (j2-j1>3 && (tmp = fun(i1,j1+1,i2,j2))<cur) {
					j1++;
					cur=tmp;
				}
				
				if ((tmp = fun(i1, j2-3, i2, j2)) < cur) {
					j1=j2-3;
					cur=tmp;
				}
				ans=min(ans, cur);
			}
		}
	}
	
	
	printf("%d\n",ans);
}


int main()
{
	int ttt;
	scanf("%d",&ttt);
	while (ttt--) {
		solve();
	}
	return 0;
}

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来源： https://blog.csdn.net/jackypigpig/article/details/120573799