hd 1017 A Mathematical Curiosity-优快云博客

本文介绍了一道关于寻找整数对(a, b)的数学题目，使得特定条件下(a² + b² + m) / (ab)为整数。文章详细阐述了问题的输入输出格式，并给出了解决该问题的示例代码。

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A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26129 Accepted Submission(s): 8275

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

Sample Output

  
   Case 1: 2
Case 2: 4
Case 3: 5

Source

East Central North America 1999, Practice

#include<stdio.h>
int main()
{
    int t,i,n,m,a,b;
    int k,c=0;
    while(scanf("%d",&t)!=EOF)
    {
     for(i=0;i<t;i++)
      {
                     k=1;
        while(scanf("%d%d",&n,&m)&&(m!=0||n!=0))
        {
        for(a=1;a<n-1;a++)
        {
         for(b=a+1;b<n;b++)
         if((a*a+b*b+m)%(a*b)==0)          c++;
         }
         printf("Case %d: %d\n",k++,c);
         c=0;
                }
    if(i!=t-1)
    printf("\n");
}
                              }
return 0;
    }

/*1.如何计算a,b的对数