HDU 1017 A Mathematical Curiosity

题目：

输入数据N ，代表N组多实例；

输入n,m代表两个数，且 a<b<n（n<100）

如果有a，b，使(a^2+b^2 +m)/(ab)为整数,（既互相取余为0）

则（a,b)为一组数；问输入的数据中，有多少组这样的数据（a，b）；

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 
     
     
     
     

想明白后是个超级大水题，但是由于手残，让b==n的情况出现了，结果就错了；

这里有个小技巧，因为b>a。故有，b=a+1；

因为a b 都小于n ，则有 a<n-1 , b<n;

这样搜索效率比直接卡100边界快得多了；

还有就是不要忘了初始化每一次实例满足组数的个数，都得从1开始；

#include <bits/stdc++.h>
using namespace std ;
int main()
{
	int a , b ,n , m; 
	int t ,k;
	int sum;
	cin>>t;
	
	while(t--)
	{
		k=1;
		while(cin>>n>>m,n+m)
		{
			sum=0;
			for(int a = 1 ; a<n-1;a++)
			{
				for(int b =a+1 ; b<n;b++)
				{
					if((a*a+b*b+m)%(a*b)==0)
					{
						sum++;
					}
				}
			}
			printf("Case %d: %d\n",k++,sum);
		}
		if(t!=0)
		printf("\n");
	}
	return 0 ;
}