杭电 1017 A Mathematical Curiosity

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19933    Accepted Submission(s): 6160

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

 

这道题的描述是有一些错误的，但是可能并不像我刚开始想的那么多错误！

问题出在哪儿呢？什么时候应该有空白行，什么时候不应该有，这点很是纠结！刚开始没看明白……

提交了3次，前两次都是PE，最后一次AC，代码如下:

 

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    int N, i, j, n, m, cnt,sum;
    scanf( "%d", &N );
    while( N-- )
    {
           cnt = 0;
           while( (scanf( "%d%d", &n, &m )!= EOF)&&(n||m) )
           {
                  cnt++;
                  sum = 0;
                  for( i = 2; i < n; i++ )
                       for( j = 1; j < i; j++ )
                            if( (i*i+j*j+m)%(i*j)==0 )
                                sum++;
                  printf( "Case %d: %d\n", cnt, sum );      
           }
           if( N != 0 )
               printf( "\n" );
    }
    
  
  //system("PAUSE");    
  return 0;
}