杭电ACM OJ 1017 A Mathematical Curiosity 题很简单，就是题目意思太让人费解了！

A Mathematical Curiosity

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 44357    Accepted Submission(s): 14219
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

1

10 1
20 3
30 4
0 0

 

[align=left]Sample Output[/align]

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

翻译：这个1我也不管他了，对理解题目意思没啥帮助。就是输入n，m。如第一例，n是10，m是1。要找0<a<b<n的数，使得(a^2+b^2 +m)/(ab)可以整除，求输入的n，m有几种结果。（具体哪坑我就不说了，可能是我个人原因）
部分代码

private void calculate() {
for (int i = 1; i < n - 1; i ++) {
for (int j = i + 1; j < n; j ++) {
if (((i + j) * (i + j) + m) % (i * j) == 0) {
System.out.println(i + " " + j);
}
}
}
}

全部代码

public class AMathematicalCuriosity1017 {
private int n;
private int m;
private AMathematicalCuriosity1017(int n, int m) {
this.n = n;
this.m = m;
}
private void calculate() {
for (int i = 1; i < n - 1; i ++) {
for (int j = i + 1; j < n; j ++) {
if (((i + j) * (i + j) + m) % (i * j) == 0) {
System.out.println(i + " " + j);
}
}
}
}
public static void main(final String[] args) throws Exception {
AMathematicalCuriosity1017 aMathematicalCuriosity1017 = new AMathematicalCuriosity1017(30, 4);
aMathematicalCuriosity1017.calculate();
}
}