A Mathematical Curiosity

바보 B원에서 전재하다
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
 
Recommend
JGShining
 
작자
Michael
문제 풀이 사고방식
이 문제는 간단한 문제입니다. 문제의 대의는 제시된 두 개의 수 n, m를 구하는 것입니다. 0#include<iostream> #include<string.h> #include<stdio.h> using namespace std; int main() { int test,i,j,n,m,t=1; scanf("%d",&test); while(test--) { getchar(); t=1; while(scanf("%d%d",&n,&m)&&n!=0||m!=0) { int sum=0; for(i=1;i<n;i++) { for(j=i+1;j<n;j++) { if((i*i+j*j+m)%(i*j)==0) { sum++; } } } printf("Case %d: %d
",t++,sum); } if(test>=1) { printf("
"); } } return 0; }