1017.A Mathematical Curiosity

最新推荐文章于 2018-08-24 11:08:18 发布

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本文探讨了一道算法题目，该题目要求计算满足特定条件的整数对(a, b)的数量。具体条件为0<a<b<n且(a²+b²+m)/(ab)为整数。通过给出的示例输入和输出，文章提供了相应的C语言实现代码。

原题描述：

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

Sample Output

Case 1: 2
Case 2: 4
Case 3: 5

忘记在完成一次输入之后换行。。。一直都是格式错误。尴尬。

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