HDU 1017 A Mathematical Curiosity【水，坑】

最新推荐文章于 2025-08-11 09:20:14 发布

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最新推荐文章于 2025-08-11 09:20:14 发布

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原文链接：http://www.cnblogs.com/ECJTUACM-873284962/p/7216283.html

本文解析了一道关于寻找整数对(a, b)的数学竞赛题，使得(a^2 + b^2 + m) / (a * b)为整数，通过双层循环遍历求解，最终给出了正确的AC代码。

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A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41995 Accepted Submission(s): 13502

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

Sample Output

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

Source

East Central North America 1999, Practice

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1017

分析：一道水题被我写的乱七八糟的，各种格式不对，首先m,n只要有一个为0就break，然后就是这个输出空行，输出格式输错了，GG！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m;
 4 int main()
 5 {
 6     int t;
 7     scanf("%d",&t);
 8     while(t--)
 9     {
10         int k=1;
11         while(scanf("%d%d",&n,&m)&&n||m)
12         {
13             int ans=0;
14             for(int i=1;i<n;i++)
15             {
16                 for(int j=i+1;j<n;j++)
17                 {
18                     if((i*i+j*j+m)%(i*j)==0)
19                         ans++;
20                 }
21             }
22            printf("Case %d: %d\n",k++,ans);
23          }
24          if(t)
25             printf("\n");
26     }
27     return 0;
28 }