GTY's math problem——大数幂比较（好伤心的精度。。。）

最新推荐文章于 2019-05-05 18:26:07 发布

A_Happy_ACMer

最新推荐文章于 2019-05-05 18:26:07 发布

阅读量548

点赞数

CC 4.0 BY-SA版权

分类专栏： =====数学=====

本文链接：https://blog.youkuaiyun.com/A_Happy_ACMer/article/details/43611055

=====数学===== 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一道关于比较两个指数表达式大小的问题，并提供了一种通过取对数的方法来解决该问题的算法实现。该算法能够处理多个测试用例，并在给定的时间和内存限制内得出正确的比较结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

GTY's math problem

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

GTY is a GodBull who will get an Au in NOI . To have more time to learn algorithm knowledge, he never does his math homework. His math teacher is very unhappy for that, but she can't do anything because GTY can always get a good mark in math exams. One day, the math teacher asked GTY to answer a question. There are four numbers on the blackboard - a,b,c,d. The math teacher wants GTY to compare ab with cd. Because GTY never does his homework, he can't figure out this problem! If GTY can't answer this question correctly, he will have to do his homework. So help him!

Input

Multi test cases (about 5000). Every case contains four integers a,b,c,d(1≤a,b,c,d≤1000)separated by spaces. Please process to the end of file.

Output

For each case , if ab>cd , print '>'. if ab<cd , print '<'. if ab=cd , print '='.

Sample Input

2 1 1 2
2 4 4 2
10 10 9 11

Sample Output

>
=
<
题意很简单，比较a的b次方和c的d次方的大小，
取对数比较就好了log(ab)=b∗log(a),注意会有精度误差。
精度误差啊，我就直接上取整了，额。。。。。。，欲哭无泪额。。。。。。
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define eps 1e-8

using namespace std;

int main()
{
    int a,b,c,d;
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        double aa = (double)b * log(a);
        double bb = (double)d * log(c);
        if(aa - bb > eps)
            printf(">\n");
        else if(aa - bb < -eps)
            printf("<\n");
        else//在一个很小的范围内，代表了很接近0，所以就近似地看成0
            printf("=\n");
    }
    return 0;
}