hdu_1017_A Mathematical Curiosity_201310280948

A Mathematical Curiosity

http://acm.hdu.edu.cn/showproblem.php?pid=1017

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22366    Accepted Submission(s): 7021

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

 

题意：

先输入一个数N然后会分N块输入，每块输入每次2个数，n,m，知道n=m=0时结束，当a和b满足0<a<b<n且使(a^2+b^2 +m)/(ab) 的值为整数时，那么这对a和b就是一组

            解 输出每次n和m对应的所有解的个数，但是要注意在每一块的输出中间有空行。

  第一个输入的N的作用是 是有几个CASE的意思

代码：

#include <stdio.h>
#include <math.h>

int main()
{
    int N;
    scanf("%d",&N);
    while(N--){
    int k=1;
    int m,n; 
    while(scanf("%d %d",&n,&m),n||m)
    {
        int i,j,num=0;
        double t;
        for(i=1;i<n;i++)
        for(j=i+1;j<n;j++)
        {
            t = (i*i+j*j+m)*1.0/(i*j);
            if(fabs(t-(int)t)<0.00000001)//这两句还可以这样处理   (i*i+j*j+m)*1.0%(i*j)==0 
            num++;
        }
        printf("Case %d: %d\n",k++,num);        
    }
    if(N)
    printf("\n");
}
    return 0;
}

弄清题意