BZOJ 3505 数三角形

枚举i,j相当于枚举两点且确定里面还有整点。好巧啊。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1050
using namespace std;
long long n,m,c[maxn*maxn][4];
long long gcd(long long a,long long b)
{
    if (b==0) return a;
    return gcd(b,a%b);
}
void get_table()
{
    c[0][0]=1;
    for (long long i=1;i<=(n+1)*(m+1);i++)
    {
        c[i][0]=1;
        for (long long j=1;j<=3;j++)
            c[i][j]=c[i-1][j]+c[i-1][j-1];
    }
}
long long get_ans()
{
    long long ret=0;
    for (long long i=1;i<=n;i++)
        for (long long j=1;j<=m;j++)
        {
            long long d=gcd(i,j);
            if (d!=1) ret+=(d-1)*(n-i+1)*(m-j+1);
        }
    return 2*ret;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    get_table();
    printf("%lld\n",c[(n+1)*(m+1)][3]-(m+1)*c[n+1][3]-(n+1)*c[m+1][3]-get_ans());
    return 0;
}