HDU 1017 A Mathematical Curiosity

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A Mathematical Curiosity

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

Total Submission(s): 36360 Accepted Submission(s): 11605

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

110 120 330 40 0

Sample Output

Case 1: 2Case 2: 4Case 3: 5题目大意是先输入一个数N然后会分N块输入，每块每次输入2个数，n,m，n=m=0时结束，当a和b满足0<a<b<n且使(a^2+b^2 +m)/(ab) 的值为
整数时，那么这对a和b就是一组，输出这样的组数，一行输入，跟着一样输出。
#include<iostream>using namespace std;int main(){int T;//记录每一块cin>>T;while(T--){int n,m;int flag=0;                   //标记第几个 case while(cin>>n>>m&&(n!=0))    //只需要判断n是否为零就可以了 {int count=0;                  //记录有多少种情况 flag++; for(int a=1;a<n;a++) {for(int b=a+1;b<n;b++){if((a*a+b*b+m)%(a*b)==0)count++; }}cout<<"Case "<<flag<<": "<<count<<endl;}if(T)cout<<endl;            //刚开始少了这一部没有通过，显示  Presentation Error 对输出要求太严格了  }  return 0; }

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