ACM HDU 1017 A Mathematical Curiosity

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原文链接：http://www.cnblogs.com/liaoguifa/archive/2012/10/06/2713095.html

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本文探讨了一个数学问题，即在给定整数n和m的条件下，找出所有满足条件的整数对(a, b)，使得0 < a < b < n且(a² + b² + m) / (ab)为整数。文章通过实例演示了解题过程，并提供了源代码。

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A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17031 Accepted Submission(s): 5306

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
  

Source

East Central North America 1999, Practice

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#include<stdio.h>
int main()
{
    int n, m, times, count = 0, i, j, flag = 1;
    scanf("%d", &times);
    while(times--)
    {
        while(1)
        {
            scanf("%d%d", &n, &m);
            if(!n && !m) break;
            for(i=1; i<n; ++i)
            for(j=1; j<i; ++j)
            if((i*i+j*j+m)%(i*j) == 0) count++;
            printf("Case %d: %d\n", flag++, count);
            count = 0;
        }
        flag = 1;
        if(times != 0) printf("\n");
    }
    
return 0;
}