CodeForces 602A Two Bases(简单题,比较两个不同进制数的大小)——Codeforces Beta Round #333 (Div. 2)

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：签到题，输入格式如下：

n1 m1

a1 a2 a3 …… an1

n2 m1

b1 b2 b3 …… bn2

n1为第一个数的位数，m1为第一个数是几进制数

n2和m2的意思同上

我们要做的是比较两个数的大小关系

解题思路：此题暴力解就可以了，要比较大小，当然得保持进制一致，最简单的莫过于都转化为十进制数

基本公式：sum=sum*m+ai

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-4
using namespace std;
const int N = 40;
const int M = 100000;
const int inf = 100000;
const int mod = 2009;
int main()
{
    int n1,m1,n2,m2,x,i;
    __int64 sum1,sum2;
    sum1=sum2=0;
    scanf("%d%d",&n1,&m1);
    for(i=0;i<n1;i++)
    {
        scanf("%d",&x);
        sum1=sum1*m1+x;
    }
    scanf("%d%d",&n2,&m2);
    for(i=0;i<n2;i++)
    {
        scanf("%d",&x);
        sum2=sum2*m2+x;
    }
    if(sum1>sum2)
        puts(">");
    else if(sum1<sum2)
        puts("<");
    else
        puts("=");
    return 0;
}

菜鸟成长记