TOJ 1154. A Mathematical Curiosity

一直wa，后来才发现是判断的问题，不能当m==0且n==0时才break，而是只有n为0时才break

Given two integers n and m, count the number of pairs of integers (a,b) such that 0 <a < b < n and (a²+b²+m)/(ab) is an integer.

Input

You will be given a number of cases in the input. Each case is specified by a line containing the integersn 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

10 120 330 40 0

Sample Output

Case 1: 2Case 2: 4Case 3: 5

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