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, numbersX and Y realised that they have different bases, which complicated their relations.

You're given a number X represented in baseb_x and a numberY 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), wheren is the number of digits in the b_x-based representation ofX.

The second line contains n space-separated integersx₁, x₂, ..., x_n (0 ≤ x_i < b_x) — the digits ofX. 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 integersm and b_y (1 ≤ m ≤ 10,2 ≤ b_y ≤ 40,b_x ≠ b_y), wherem is the number of digits in the b_y-based representation ofY, and the fourth line contains m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y) — the digits ofY.

There will be no leading zeroes. Both X andY 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 21 0 1 1 1 12 104 7

Output

=

Input

3 31 0 22 52 4

Output

<

Input

7 1615 15 4 0 0 7 107 94 8 0 3 1 5 0

Output

>

Note

In the first sample, X = 101111₂ = 47₁₀ = Y.

In the second sample, X = 102₃ = 21₅ andY = 24₅ = 112₃, thusX < Y.

In the third sample, andY = 4803150₉. We may notice thatX starts with much larger digits and b_x is much larger thanb_y, soX 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;}

。