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.
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
10 1 20 3 30 4 0 0
Sample Output
Case 1: 2 Case 2: 4 Case 3: 5
数字不大，直接暴力过就行
#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1e6 + 5;
int dp[maxn],a[maxn];
int main() {
int n,m,t = 0;
while(scanf("%d%d",&n,&m) != EOF )
{
if(n == 0 && m == 0)
break;
t++;
int count = 0;
for(int i = 1; i < n; i++)
{
for(int j = i + 1; j < n ;j++)
{
double x = (i*i + j*j + m)/(i*j*1.0);
//printf("%.1lf\n",x);
if(x - (int)x == 0)
count ++;
}
}
cout<<"Case "<<t<<": "<<count<<endl;
}
return 0;
}