A Mathematical Curiosity 分类： HDU 2015-06-25 21:27 11人阅读 评论(0) 收藏

A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31177 Accepted Submission(s): 9988

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
题意简单，就是找合适的a,b满足条件
就是注意输出

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int n,m;
    int h;
    int w;
    while(~scanf("%d",&w))
    {
        for(int k=0; k<w; k++)
        {
            if(k)
                cout<<endl;
            h=1;
            while(scanf("%d %d",&n,&m)&&(n||m))
            {
                int sum=0;
                for(int i=1; i<n; i++)
                {
                    for(int j=i+1; j<n; j++)
                    {
                        if((i*i+j*j+m)%(i*j)==0)
                        {
                            sum++;
                        }
                    }
                }
                printf("Case %d: %d\n",h++,sum);
            }
        }
    }
    return 0;
}

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