bzoj3505 Cqoi2014—数三角形

Posted 2020-08-24 MashiroSky

http://www.lydsy.com/JudgeOnline/problem.php?id=3505 (题目链接)

题意

　　给定一个n*m的网格，请计算三点都在格点上的三角形共有多少个。

Solution

　　$${ans=平面中选三个点的方案数-三点共线的方案数}$$

　　$${ans=C_{(n+1)*(m+1)}^{3}-(n+1)*C_{m+1}^{3}-(m+1)*C_{n+1}^{3}-斜的三点共线的方案数}$$

　　斜的三点共线方案数不会求。。左转题解：http://blog.csdn.net/zhb1997/article/details/38474795

细节

　　LL

代码

// bzoj3505
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define inf 10000000
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

int n,m;
LL c[2000010][4];

int gcd(int a,int b) {
	return b==0 ? a : gcd(b,a%b);
}
int main() {
	scanf("%d%d",&n,&m);
	for (int i=0;i<=(n+1)*(m+1);i++) c[i][0]=1;
	for (int i=1;i<=(n+1)*(m+1);i++)
		for (int j=1;j<=min(3,i);j++) c[i][j]=c[i-1][j-1]+c[i-1][j];
	LL ans=c[(n+1)*(m+1)][3]-(n+1)*c[m+1][3]-(m+1)*c[n+1][3];
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++) {
			LL x=gcd(i,j)+1;
			if (x>2) ans-=(x-2)*2*(n-i+1)*(m-j+1);
		}
	printf("%lld",ans);
	return 0;
}