ZOJ 3647 Gao the Grid （n*m 中格点三角形个数）

Gao the Grid

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

此题就是求格点中三角形的个数。

就是找出三点不共线的个数。

n*m的矩形中有(n+1)*(m+1)个格点。

选出三个点的总个数为：C((n+1)*(m+1),3).

减掉共线的情况就是答案了。

首先是水平和垂直共线的情况：C(n+1,3)*(m+1)+C(m+1,3)*(n+1);

然后斜的共线的情况就是枚举矩形。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

int gcd(int a,int b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
long long Com(int n,int r)
{
    if(n<r)return 0;//这个一定要
    if(n-r<r)r=n-r;
    int i,j;
    long long ret=1;
    for(i=0,j=1;i<r;i++)
    {
        ret*=(n-i);
        for(;j<=r&&ret%j==0;j++)ret/=j;
    }
    return ret;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        long long  ans=Com((n+1)*(m+1),3);//选三个点的所有组合数
        for(int i=2;i<=n;i++)
          for(int j=2;j<=m;j++)
          {
              ans-=(long long)(gcd(i,j)-1)*(n-i+1)*(m-j+1)*2;
          }
        ans-=Com(n+1,3)*(m+1);
        ans-=Com(m+1,3)*(n+1);
        printf("%lld\n",ans);//ZOJ用lld,不能用I64d
    }
    return 0;
}