[ACM]A Mathematical Curiosity

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

East Central North America 1999, Practice

解题思路：枚举。

代码：

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int N,n,m,Case,count;
    int Ni,a,b;
    cin>>N;
    for(Ni=0;Ni<N;Ni++)
    {
        Case=1;
        while(cin>>n>>m&&(n||m))//这个条件要特别注意
        {
            count=0;
            for(a=1;a<n;++a)
                for(b=a+1;b<n;++b)
            {
                //判断一个数是否为整数的第一种方法
                if((a*a+b*b+m)%(a*b)==0)
                    count++;

                //第二种方法

                /*float temp;
                temp=(a*a+b*b+m)/(a*b*10.0)*10.0;
                if(temp-int(temp)==0)
                    count++;*/

                //第三种方法

                /*float temp;
                temp=(a*a+b*b+m)/(a*b*10.0)*10.0;
                if(floor(temp+0.5)==temp)
                    count++;*/
            }
            cout<<"Case "<<Case<<": "<<count<<endl;
            Case++;
        }
        if(Ni<N-1)
            cout<<endl;
    }
    return 0;
}

误区：这个题目看起来很简单，但有一个小细节也会让你提交N遍也是Wrong Answer，就是那个while(cin>>n>>m&&(n||m))里面的条件，是&&(n||m) 而不是&&n&&m,正是因为这个小条件弄错了，郁闷的我快抓破了头皮,前者的意思是 n和m不能同时为0，m可以为0，后者的意思是n和m都不能等于0，而题目的意思是n不能等于0，但m可以等于0.  哎！教训啊！

运行截图：