CF--333--div2--A.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 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
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 bx is much larger than by, so X is clearly larger than Y.


题意:给出两个不同进制的数,判断两个数是否相等。

思路:直接将每个不同进制的数转成10进制再统一进行比较即可。

任意进制 x 转化为 10 进制,将 x 进制的数 v[n] 从低位到高位依次用 x*v[i]^(i),下标 i 从0开始。

code:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;
ll trans(ll v, int k, int n) {
	while(k--)
		v*=n;
	return v;
}
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	int n, m, x, y, i, j;
	ll p=0, q=0, v;
	scanf("%d %d", &n, &x);
	for(i=0;i<n;i++){
		scanf("%lld", &v);
		if(v)
			p+=trans(v, n-i-1, x);//将x进制的数v转化为10进制
	}
	scanf("%d %d", &m, &y);
	for(i=0;i<m;i++){
		scanf("%lld", &v);
		q+=trans(v, m-i-1, y);
	}
	if(p==q)
		printf("=\n");
	else if(p>q)
		printf(">\n");
	else
		printf("<\n");
	return 0;
}

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