cf 602 A(进制转换)

A. Two Bases

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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,

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;
}