[catch]--A Mathematical Curiosity

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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【C++】

/*
* [程序设计实践与提高]
* 题目：A Mathematical Curiosity 
* 注意输出的格式：每输入一组数据就输出一组数据	
* 时间：2013年9月12日
*/

#include <iostream>

using namespace std;

int main()
{
	int N;

	cin >> N;
//	cout << endl;//换行

	//N个输入块
	while (N--)
	{
		int n;
		int m;
		int count1 = 0;//记录输入的数据组数，用来输出case序列

		while (cin >> n >> m)
		{
			if (0 == n && 0 == m)
			{
				break;
			}
			else
			{
				int count2 = 0;//满足条件的个数
				int i;
				int j;

				for (i = 1; i < n; ++i)
				{
					for (j = i+1; j < n; ++j)
					{
						if (i > 0 && i < j && j < n && (0 == (i*i+j*j+m)%(i*j)))
							++count2;
					}				
				}	
				cout << "Case " << ++count1 << ": " << count2 << endl;
			}
		}	

		if (N)
			cout << endl;
	}

	return 0;
}

 

【C】

#include <stdio.h>

int main(void)
{
	int N;
	scanf("%d", &N);

	while (N--)
	{
		int n;
		int m;
		int count1 = 0;
		while (scanf("%d %d",&n,&m))
		{
			if (n == 0 && m == 0)
				break;
			int count2 = 0;
			int i;
			int j;

			for (i = 1; i < n; ++i)
			{
				for (j = i+1; j < n; ++j)
				{
					if (i > 0 && i < j && j < n && (0 == (i*i+j*j+m)%(i*j)))
						++count2;
				}				
			}	
			printf("Case %d: %d\n", ++count1, count2);
		}	

		if (N)
			printf("\n");
	}

	return 0;
}

 

【结果图片】