A Mathematical Curiosity（坑水题）

最新推荐文章于 2020-03-15 20:52:25 发布

叶孤心丶

最新推荐文章于 2020-03-15 20:52:25 发布

阅读量1.2k

点赞数

CC 4.0 BY-SA版权

分类专栏： ---- HDU ----

本文链接：https://blog.youkuaiyun.com/yeguxin/article/details/46757221

---- HDU ---- 专栏收录该内容

102 篇文章

订阅专栏

本文探讨了一个有趣的数学现象，被称为‘坑水题’，引领读者深入理解其背后的数学原理和奇妙之处。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A Mathematical Curiosity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31228 Accepted Submission(s): 10004

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

Recommend

JGShining | We have carefully selected several similar problems for you: 1018 1005 1021 1019 1032

题目没有算法，直接枚举就可以，不过这个输入的要求有点奇葩，以前没见过这样的。一直没明白过来到底是什么意思，写完直接交了直接WA，斗争很久就是交不上，看了一个前辈的一句话，才知道是这样....

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int k = 0;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            if(n == 0 && m == 0)
            {
                break;
            }
            int cnt = 0;
            for(int i=1; i<n; i++)
            {
                for(int j=i+1; j<n; j++)
                {
                    if((i*i + j*j + m)%(i*j) == 0)
                    {
                        cnt++;
                    }
                }
            }
            printf("Case %d: %d\n",++k,cnt);

        }
        if(T)
        {
            printf("\n");
        }
    }

    return 0;
}