hdu1017A Mathematical Curiosity

A Mathematical Curiosity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 48317    Accepted Submission(s): 15520 http://acm.hdu.edu.cn/showproblem.php?pid=1017   Problem Description Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer. This problem contains multiple test cases! The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks. The output format consists of N output blocks. There is a blank line between output blocks.   Input You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.   Output For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.   Sample Input 1 10 1 20 3 30 4 0 0   Sample Output Case 1: 2 Case 2: 4 Case 3: 5

题意：输入N个输入块，块间有空格。每个块中，多组样例，输入n，m。当n=0时，此输入块结束。输出满足(a^2+b^2 +m)/(ab)的（a,b）种数(0 < a < b < n)。

题目很简单，就是原题说n=m=0时输出结束，但这个题只有n=0时输出结束才对。ayixi~~

代码：

#include<iostream>
#include<cstring>
#include<cmath>
#include<iomanip>
#include<algorithm>
#include<cstdio>
#define inf 0x3f3f3f3f
using namespace std;

int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n,m;
        int h=1;
        while(cin>>n>>m&&n!=0)
        {
            int sum=0;
            for(int b=2;b<n;b++)
            {
                for(int a=1;a<b;a++)
                {
                    if((a*a+b*b+m)%(a*b)==0)
                        sum++;
                }
            }
            cout<<"Case "<<h++<<": "<<sum<<endl;
        }
        if(N!=0)
            cout<<endl;
    }
    return 0;
}