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
题目大意：先输入一个数N然后会分N块输入，每块中会每次输入2个数，n,m，n=m=0时结束本块的输入，当a和b满足0

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<string>#include<cmath>#include<queue>#include<vector>#define N 201using namespace std;int main(){    int x,n,m,i,j,s,c;    scanf("%d", &x);    while(x--)    {        c=0;        while(scanf("%d %d", &n,&m)!=EOF&&n!=0)        {            s=0;            for(i=1; i<n; i++){                for(j=i+1; j<n; j++){                    if((i*i+j*j+m)%(i*j)==0)                        s++;                }            }            printf("Case %d: %d\n", ++c, s);        }        if(x)            printf("\n");    }    return 0;}

再因为换行错了我就可以去死一会了。。。