ZOJ 3647 Gao the Grid（居然是暴力）

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 integers n and m (1 ≤ n, m ≤ 1000).

Output

For each case, output the total number of triangle.

Sample Input

1 1
2 2

Sample Output

4
76

hint

hint for 2nd case: C(9, 3) - 8 = 76

題目意思 ： 給你矩形長度，求在裡面取3個點，問可以組成三角形的個數？

一共有（n+1）*（m+1）個點，去3個

然後減去同行取3個，同列取3個；

最後減去左斜和右斜的，這種情況居然是枚舉三角形的長高！！！！！

#include
#include
#include
#include
#include
#include
#include
#include

#define eps 1e-8
using namespace std;
#define N 10005

typedef 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