HDU 1017 A Mathematical Curiosity【看懂题意+穷举法】

//2014.10.17    01:19

//题意：

//先输入一个数N，然后分块输入，每块输入每次2个数，n，m，直到n,m同一时候为零时 

//结束，当a和b满足题目要求时那么这对a和b就是一组

//注意：

//每一块的输出中间有一个回车

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 27177    Accepted Submission(s): 8651

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

field=problem&key=East+Central+North+America+1999%2C+Practice&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">East Central North America 1999, Practice

 

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