标签：acm算法   c   math.h   printf   amp   

GTY‘s math problem

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!

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.

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

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

>
=
<

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);

OK

可以写代码了。

控制到12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;
}

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

标签：acm算法   c   math.h   printf   amp   

原文地址：http://blog.csdn.net/sky_miange/article/details/43611547