Codeforces 602A Two Bases

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

直接把每个数都转换成同一个进制，然后比较就行，我是都转换成了十进制

但是！！！不能使用pow函数，就是说不能求幂，必须用秦九韶算法（霍纳法则）才能AC，否则会TLE

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
	long long num[2] = { 0 };
	long long n, b;
	for (int e = 0; e < 2; ++e)
	{
		long long temp;
		scanf("%I64d%I64d", &n, &b);
		scanf("%I64d", &num[e]);
		for (int i = 0; i < n - 1; ++i)
		{
			scanf("%I64d", &temp);
			num[e] = num[e] * b + temp;
		}
	}
	
	if (num[0] < num[1])
		printf("<");
	else if (num[0]>num[1])
		printf(">");
	else if (num[0] == num[1])
		printf("=");
	return 0;
}

。