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最新推荐文章于 2018-09-27 16:17:59 发布

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本文介绍了一种比较两个不同基数表示数大小的算法，并通过具体样例展示了如何实现这一过程。输入包括两个数的不同基数及其对应的数字序列，输出为两数之间的大小关系。

After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers X and Y realised that they have different bases, which complicated their relations.

You're given a number X represented in base bx and a number Y represented in base by. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and bx (1 ≤ n ≤ 10, 2 ≤ bx ≤ 40), where n is the number of digits in the bx-based representation of X.

The second line contains n space-separated integers x1, x2, ..., xn (0 ≤ xi < bx) — the digits of X. They are given in the order from the most significant digit to the least significant one.

The following two lines describe Y in the same way: the third line contains two space-separated integers m and by (1 ≤ m ≤ 10,2 ≤ by ≤ 40, bx ≠ by), where m is the number of digits in the by-based representation of Y, and the fourth line contains m space-separated integers y1, y2, ..., ym (0 ≤ yi < by) — the digits of Y.

There will be no leading zeroes. Both X and Y will be positive. All digits of both numbers are given in the standard decimal numeral system.

Output

Output a single character (quotes for clarity):

'<' if X < Y
'>' if X > Y
'=' if X = Y

Sample test(s)

input
6 2
1 0 1 1 1 1
2 10
4 7

output
=

input
3 3
1 0 2
2 5
2 4

output
<

input
7 16
15 15 4 0 0 7 10
7 9
4 8 0 3 1 5 0

output
>

Note

In the first sample, X = 1011112 = 4710 = Y.

In the second sample, X = 1023 = 215 and Y = 245 = 1123, thus X < Y.

In the third sample, $\text{[math]}$ and Y = 48031509. We may notice that X starts with much larger digits and bx is much larger than by, so X is clearly larger than Y.

#include <stdio.h>
using namespace std;

__int64 a[20],b[20];
__int64 res,ans;

int main()
{
    int n,m;
    res=ans=0;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        scanf("%I64d",a+i);
    __int64 aa,bb;
    aa=1;
    for(int i=n-1;i>=0;i--)
    {
        res+=a[i]*aa;
        aa*=m;
    }
 //   printf("%I64d\n",res);
    int x,y;
    scanf("%d%d",&x,&y);
    for(int i=0;i<x;i++)
        scanf("%I64d",b+i);
    bb=1;
    for(int i=x-1;i>=0;i--)
    {
        ans+=b[i]*bb;
        bb*=y;
    }
 //   printf("%I64d\n",ans);
    if(res>ans)
        printf(">\n");
    else if(res==ans)
        printf("=\n");
    else
        printf("<\n");


    return 0;
}