Codeforces Round #333 (Div. 2) A. Two Bases （进制转换比较大小）

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.

题意：给你一个n和bx，n表示转换为bx进制后的位数，然后输入n个数，然后给你一个m和by，m表示转换为by进制后的位数，然后输入m个数，比较两个数的大小

思路：直接全部都转换为10进制，然后比较大小就行了

总结：水题，看懂样例就能写了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 200010
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
ll fun(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b%2)
		ans=ans*a;
		a=a*a;
		b/=2;
	}
	return ans;
}
int main()
{
	ll n,m,x,y,i,a; 
	while(scanf("%I64d%I64d",&n,&x)!=EOF)
	{
		ll num1=0;
		for(i=n-1;i>=0;i--)
		{
			scanf("%I64d",&a);
			num1+=fun(x,i)*a;
		}
		scanf("%I64d%I64d",&m,&y);
		ll num2=0;
		for(i=m-1;i>=0;i--)
		{
			scanf("%I64d",&a);
			num2+=fun(y,i)*a;
		}
		if(num1>num2)
		printf(">\n");
		else if(num1==num2)
		printf("=\n");
		else
		printf("<\n");
	}
	return 0;
}