HDU-OJ-1017 A Mathematical Curiosity-整数对

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本文探讨了一个数学问题，即在给定整数n和m的条件下，找出所有满足特定数学表达式的整数对(a, b)的数量。通过详细解释输入输出格式、案例分析和解决方案，本文旨在帮助读者理解并解决此类问题。

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32002 Accepted Submission(s): 10254

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

Source

East Central North America 1999, Practice

Recommend

JGShining

#include<iostream>
using namespace std;
int main()
{
    int m,n,t,j,i,k,ans,count;
    cin>>t;
    for(k=0; k<t; k++)
    {
        count=0;
        if(k)
            cout<<endl;
        while(cin>>n>>m&&(m||n))
        {
            ans=0;
            for(i=1; i<n; i++)
                for(j=i+1; j<n; j++)
                {
                    if((i*i+j*j+m)%(i*j)==0)
                        ans++;
                }
            cout<<"Case "<<++count<<": "<<ans<<endl;
        }
    }
    return 0;
}

奇坑无比的一道题！！！

WA了好多次，看了好几遍题目才整明白！！！

前面那个“1”表示下面有0 0 之前的一个大组测试数据。

所以，一大组测试数据之前的一个整数表示有几个大组包含若干小组。

还要注意，输出空行。