CodeForces 602 A. Two Bases

最新推荐文章于 2020-02-09 20:52:05 发布

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本文介绍了一个简单的程序设计问题，即如何将不同基数的数字转换为十进制，并进行大小比较。通过具体的样例输入输出展示了问题解决的过程。

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题目描述：

水题，只是注意精度，还有如果你要用pow()函数来求解幂，那么注意pow()返回值为double，如果你用long long int 那么就wrong了，及时类型转换可是在大数据时还是出现数据错乱，不知道为什么。

思路：都转换成十进制然后比较就行

提后分析：

用c做十进制的输入输出格式
int %d
long int %ld
long long int %I64d
unsigned int %d
unsigned long int %ld
unsigned long long int %I64d
double 输入%lf
float 输入 %f
输出 %lf或%f都行
long double 就不知道用什么了还是用cin吧！！！

题目：

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 b_x and a number Y represented in base b_y. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and b_x (1 ≤ n ≤ 10, 2 ≤ b_x ≤ 40), where n is the number of digits in the b_x-based representation of X.

The second line contains n space-separated integers x₁, x₂, ..., x_n (0 ≤ x_i < b_x) — 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 b_y (1 ≤ m ≤ 10, 2 ≤ b_y ≤ 40, b_x ≠ b_y), where m is the number of digits in the b_y-based representation of Y, and the fourth line contains m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y) — 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

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 b_x is much larger than b_y, so X is clearly larger than Y.

代码：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <stack>
#include <queue>
#include <algorithm>
#include <math.h>

using namespace std;

 long double x, ax, y, by;
 long double DX, DY;//存储转换后十进制的数据

 long double toDecimal( long double a, long double ax)
{
     long double temp = 0;
     long double d = 0;
    for( int i=(int)a; i>0; i--)
    {
        cin>> temp;
        d += temp * pow(ax, i-1);
    }
    return d;
}

int main()
{
    while( cin>> x >> ax )
    {
        long double x1 = toDecimal(x, ax);
        cin>> y >> by;
        long double y1 = toDecimal(y, by);
        if(x1 > y1)
            cout<<">"<<endl;
        else
            if(x1 == y1)
            cout<<"="<<endl;
        else
            cout<<"<"<<endl;
    }
    return 0;
}