Codeforces--602A--Two Bases(水)
Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
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 Input
6 2
1 0 1 1 1 1
2 10
4 7
=
3 3
1 0 2
2 5
2 4
<
7 16
15 15 4 0 0 7 10
7 9
4 8 0 3 1 5 0
>
Hint
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.
Source
#include<stdio.h>
#include<string.h>
int main()
{
long long A=0,B=0;
int n,b,a;
scanf("%d%d",&n,&b);
for(int i=0;i<n;i++)
{
scanf("%d",&a);
A=A*b+a;
}
scanf("%d%d",&n,&b);
for(int i=0;i<n;i++)
{
scanf("%d",&a);
B=B*b+a;
}
if(A>B)
printf(">\n");
if(A==B)
printf("=\n");
if(A<B)
printf("<\n");
return 0;
}
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