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

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
  
   
#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