ZOJ 3647 Gao the Grid（居然是暴力）

A n * m grid as follow:

Count the number of triangles, three of whose vertice must be grid-points.
Note that the three vertice of the triangle must not be in a line(the right picture is not a triangle).

Input

The input consists of several cases. Each case consists of two positive integersn and m (1 ≤ n, m ≤ 1000).

Output

For each case, output the total number of triangle.

Sample Input

1 12 2

Sample Output

476

hint

hint for 2nd case: C(9, 3) - 8 = 76 

题目意思 ： 给你矩形长度，求在里面取3个点，问可以组成三角形的个数？

一共有（n+1）*（m+1）个点，去3个

然后减去同行取3个，同列取3个；

最后减去左斜和右斜的，这种情况居然是枚举三角形的长高！！！！！

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#define eps 1e-8using namespace std;#define N 10005typedef long long ll;ll L(ll x){return (ll)(x*(x-1)*(x-2)/6);}ll gcd(ll n,ll m){if(m==0)  return n;return gcd(m,n%m);}int main(){ll i,j,n,m;while(~scanf("%lld%lld",&n,&m)){   n++;    m++;ll ans=L(n*m)-L(n)*m-L(m)*n;for(i=2;i<n;i++)for(j=2;j<m;j++){ll temp=gcd(i,j)-1;temp=temp*(n-i)*(m-j)*2;ans-=temp; }       printf("%lld\n",ans);}    return 0;}