1、链接：

2、题目：

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

3、解题分析：

根据题意知：求在（a， b）区间内（a^2 + b^2 +ｍ）／ａ＊ｂ的结果是一个整形数的个数，即（a^2 + b^2 +ｍ）%（ａ＊ｂ）==０暴力求解即可

注意：注意输出格式每一大组之间有一个换行

４、代码

#include
     
      using namespace std;int main(){    int N;    scanf("%d",&N);    int n,m;    while(N--)    {        int cas = 1;        while(~scanf("%d%d",&n,&m)){            int count = 0;            if(n == 0 && m ==0 )                break;            for(int i = 1; i < n; i++)            {                for(int j = i+1; j< n; j++)                {                    if((i*i + j*j + m)%(i*j) == 0)                    {                        count++;                    }                }            }            printf("Case %d: %d\n",cas++,count);        }        if(N)            printf("\n");    }    return 0;}/*110 120 330 40 0*/