A Mathematical Curiosity
浏览数：24 /
时间：2015年06月11日
A Mathematical Curiosity
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 87 Accepted Submission(s) : 23
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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.
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
#include<iostream>
using namespace std;
int main()
{
int a,b,m,n,k,j,i,ans,num;
cin>>k;
for(b=0;b<k;b++)
{ num=0;
if(b)
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 "<<++num<<": "<<ans<<endl;
}
}
return 0;
}郑重声明：本站内容如果来自互联网及其他传播媒体，其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享，并不代表本站赞同其观点和对其真实性负责，也不构成任何其他建议。
