cf 602 A(进制转换)

A. Two Bases

time limit per test

memory limit per test

input

output

X and Y

X represented in base bx and a number Y represented in base by. Compare those two numbers.

Input

n and bx (1 ≤ n ≤ 10, 2 ≤ bx), where n is the number of digits in the bx-based representation of X.

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.

Y in the same way: the third line contains two space-separated integers m and by (1 ≤ m ≤ 10,2 ≤ by, 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.

X and Y

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

X = 1011112 = 4710 = Y.

X = 1023 = 215 and Y = 245 = 1123, thus X < Y.

In the third sample, 

 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;
}