HDU5170GTY's math problem

GTY's math problem

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1845    Accepted Submission(s): 694

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

>
=
<

 

看不懂题目的意思的小伙伴可以去看看BEST CODE 里面中文版的：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=567&pid=1001

其实也是一道水题恩：

咋一看，要求a^b    c^d，数据量体较大，正常情况下去算，肯定是不能过的，用快速幂和模拟大数问题去处理也是不可以实现的，肯定会超时；那么我们可以想到对其取对数

那么题目就变得很简单了：

考虑两种情况：

         a和c是否为 1 ，在合法数据里面他们是可以为 1 的，当他们两为 1 的时候那么无论 b和d 为多少他们都是相等的，就不要考虑取对数了；

         还有一种就是取对数，正常比较，当他们取对数后的差值不超过10e-12 时我们就认为这两个数时相等的；

那么就可以写代码了。

给出AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
	int a, b, c, d;
	double x,y;
	while (~scanf("%d%d%d%d",&a,&b,&c,&d))
	{
		if (a == 1 && c == 1)
			printf("=\n");
		else
		{
			x = b*1.0*log(a*1.0);
			y = d*1.0*log(c*1.0);
			if (abs(x - y) < 10e-12)
				printf("=\n");
			else if (x >y)
				printf(">\n");
			else if (x < y )
				printf("<\n");
		}
	}
	return 0;
}