传送门：
http://acm.hdu.edu.cn/showproblem.php?pid=1017
A Mathematical Curiosity
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47658 Accepted Submission(s): 15285
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
Source
分析：
直接暴力即可
code：
#include<bits/stdc++.h> using namespace std; typedef long long LL; #define max_v 105 int main() { int t; int n,m; scanf("%d",&t); while(t--) { int k=1; while(~scanf("%d %d",&n,&m)) { if(n==0&&m==0) break; int c=0; //直接暴力即可 for(int i=1; i<n-1; i++) { for(int j=i+1; j<n; j++) { if((i*i+j*j+m)%(i*j)==0) c++; } } printf("Case %d: %d\n",k++,c); } if(t!=0) printf("\n"); } return 0; }