Codeforces--602A--Two Bases(水)

Two Bases

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit
Status

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
b_x and a number
Y represented in base
b_y. Compare those two numbers.

Input

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

The second line contains n space-separated integers
x₁, x₂, ..., x_n (0 ≤ x_i < b_x)
— 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 b_y (1 ≤ m ≤ 10,
2 ≤ b_y ≤ 40,
b_x ≠ b_y), where
m is the number of digits in the
b_y-based representation of
Y, and the fourth line contains
m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y)
— 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

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

>

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
b_x is much larger than
b_y, so
X is clearly larger than Y.

Source

Codeforces Round #333 (Div. 2)

#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;
}