HDU 5170 GTY's math problem (bsst code #29 1001)

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本文介绍了一道关于比较两个幂次运算结果大小的问题，并提供了一种利用对数进行比较的有效解决方案。通过控制精度到小数点后十二位，确保了比较的准确性。

GTY's math problem

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 146 Accepted Submission(s): 80

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

Sample Output

  
   >
=
<

Source

BestCoder Round #29

这个数据一看就知道不能用普通的方法。

1000的1000次方。

C语言无法表达的。

可以想到用对数。

比较log(a^b) 和log(c^d)的大小

而log(a^b)=b*log(a);

log(c^d)=d*log(c);

可以写代码了。

控制到小数点后十二位就可以认为它们相等了

#include <stdio.h>
#include <math.h>
#define eps 1e-12  //10的负十二次方
int main()
{
	double a,b,c,d;
	while(scanf("%lf%lf%lf%lf",&a,&b,&c,&d)!=EOF)
	{
		double s,s1;
		s=b*log(a);
		s1=d*log(c);
		if(a==1 &&c==1)  //等于1肯定都是相等的
		{
			printf("=\n");
			continue;
		}  
		if(fabs(s-s1)<eps)     //对数计算后，相差只要不超过10的负十二次方，就可以认为它们相等。
			printf("=\n"); 
		else if(s>s1)
			printf(">\n");
		else  if(s<s1)
			printf("<\n");
	}
	return 0;
}