XDOJ_1031 _水

http://acm.xidian.edu.cn/problem.php?id=1031

简单找规律，a[i][j] = (i+j)/gcd(i,j)。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

int n,m;

int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int ans = 0;
        for(int i = 1;i <= n;i++)
        {
            for(int j = 1;j <= m;j++)
            {
                ans += (i+j)/gcd(i,j);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}