[HDU - 5170GTY's math problem 数的精度类

题目描述

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

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

解题思路

对于a^b ,这类数太大了，总的来说这是一个数的精度相关的题目，我们把数据的两边取对数，就可以解决这个问题
如果两个数之间相差的的数小于10 的负12次方（怎样都会有误差），就可以认定两个数相等，对于这里的用到的函数fabs是针对浮点型数据取绝对值的函数。

AC代码：

#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 m, n;
    m = b*log(a);
    n = d*log(c);
    if (a == 1 && c == 1)  //等于1肯定都是相等的
    {
        printf("=\n");
        continue;
    }
    if (fabs(m - n)<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 (m>n)
        printf(">\n");
    else  if (m<n)
        printf("<\n");
}
return 0;
}

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