Codeforces Round #333 (Div. 2) A. Two Bases

最新推荐文章于 2018-09-27 16:17:59 发布

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最新推荐文章于 2018-09-27 16:17:59 发布

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分类专栏： ACM 模拟 Codeforces 文章标签： ACM codeforces 模拟

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本文介绍了一种比较两个不同基数表示的数值大小的方法。输入包括两组数字及其对应的基数，输出则是这两个数值之间的大小关系。文章通过具体示例说明了如何进行转换及比较。

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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 bxis much larger than by, so X is clearly larger than Y.

结题报告：Codeforces Round #333 — editorial

#include
  
   
#include
   
    
#include
    
     
#include
     
      
#include
      
using namespace std;
const int N=100005;
int a[N];
int b[N];
__int64 sum=0,ans=0,kk;
int main()
{
    int n,m,x,y;
    int i,j,k;
    kk=1;
    scanf("%d%d",&n,&x);
    for(i=0; i
       
        ans) { printf(">\n"); } else { printf("<\n"); } return 0; }