BestCoder Round #29 1001 GTY's math problem

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

>
=
<

题解思路：注意精度控制：
代码：
#include <set>
#include <map>
#include <cmath>
#include <time.h>
#include <queue>
#include <deque>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
int main()
{
    int a,b,c,d;
    double ans1,ans2;
    while(scanf("%d %d %d %d",&a,&b,&c,&d) != -1)
    {
        ans1 = 1.0 * b * log10(a);
        ans2= 1.0 * d * log10(c);
        if(fabs(ans1 - ans2) < 1e-10)
            printf("=\n");
        else if(ans1 - ans2 > 1e-10)
            printf(">\n");
        else
            printf("<\n");
    }
    return 0;
}