Codeforces 602A Two Bases 【水题】

最新推荐文章于 2018-09-09 10:55:52 发布

笑着走完自己的路

最新推荐文章于 2018-09-09 10:55:52 发布

阅读量611

点赞数

CC 4.0 BY-SA版权

分类专栏：水题 codeforces

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/50162815

codeforces 同时被 2 个专栏收录

215 篇文章

订阅专栏

水题

127 篇文章

订阅专栏

本文介绍了一种方法，用于比较两个使用不同基数表示的数的大小，通过将它们转换为十进制进行比较。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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, $\text{[math]}$ 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：给定两个不同进制数，判断大小。

水题，转化为十进制比较就可以了。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f
#define eps 1e-8
#define MAXN (1000000+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int main()
{
    int n, m, b1, b2;
    LL A = 0, B = 0;
    Ri(n); Ri(b1);
    for(int i = 0; i < n; i++)
    {
        int a; Ri(a);
        A = A * b1 + a;
    }
    Ri(m); Ri(b2);
    for(int i = 0; i < m; i++)
    {
        int a; Ri(a);
        B = B * b2 + a;
    }
    if(A > B)
        printf(">\n");
    else if(A == B)
        printf("=\n");
    else
        printf("<\n");
    return 0;
}