A Mathematical Curiosity

Problem Description

Given twointegers 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 aninteger.

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 begiven a number of cases in the input. Each case is specified by a linecontaining the integers n and m. The end of input is indicated by a case inwhich n = m = 0. You may assume that 0 < n <= 100.

Output

For eachcase, print the case number as well as the number of pairs (a,b) satisfying thegiven property. Print the output for each case on one line in the format asshown below.

Sample Input

1

 

10 1

20 3

30 4

0 0停止输入，开始输出

Sample Output

Case 1: 2

Case 2: 4

Case 3: 5

#include<iostream>
using namespace std;
int main()
{
	int a,b,m,n,N,count,Case;

	cin>>N;

	int i;
	for(i=0;i<N;++i)//可以用while(N--),但是控制每个模块之间的空行要用 if(N)【N不为0时，每个模块才空行】
	{
		Case=1;//记录每个模块中输入的组数
		while(cin>>n>>m&&(n||m))//特别注意只有n和m同时为0时满足才成立，所以（n!=0&&m!=0）是不正确的
		{
			count=0;

			for(a=1;a<n;++a)
			{
				for(b=a+1;b<n;++b)
				{
					if((a*a+b*b+m)%(a*b)==0)//计算(a*a+b*b+m)/(a*b)是否为整数
					{
						count++;//记录每组数据中使要求满足的a、b的组数
					}
				}
			}
			
			cout<<"Case "<<Case<<": "<<count<<endl;
			Case++;
		}
		if(i<N-1)
			cout<<endl;
	}
	return 0;
}