time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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
Examples
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 bx is much larger than by, so X is clearly larger than Y.

【题目链接】:http://codeforces.com/contest/602/problem/A

【题解】

把它们都转换成10进制再比较就好.
40^10不会爆LL

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)typedef pair<int,int> pii;
typedef pair<LL,LL> pll;const int MAXN = 10+5;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);int n,bx,m,by;
LL a[MAXN];int main()
{//freopen("F:\\rush.txt","r",stdin);rei(n);rei(bx);rep2(i,n-1,0)rel(a[i]);LL x = 0;LL now = 1;rep1(i,0,n-1){x += now*a[i];now = now * bx;}rei(m);rei(by);rep2(i,m-1,0)rel(a[i]);LL y = 0;now = 1;rep1(i,0,m-1){y += now*a[i];now = now * by;}if (x==y)putchar('=');elseif (x < y)putchar('<');elseputchar('>');return 0;
}