1010 Radix (25分)

最新推荐文章于 2022-07-03 13:29:28 发布

原创最新推荐文章于 2022-07-03 13:29:28 发布 · 166 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

126 篇文章

订阅专栏

本文介绍了一个算法，用于解决两个数在不同进制下相等的问题。通过将其中一个数从其原始进制转换为十进制，然后搜索另一个数的进制，使得两数相等。文章详细描述了算法的实现过程，包括如何从任意进制转换到十进制，如何找到可能的最小进制，以及如何进行二分查找确定目标进制。

Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:


N1 N2 tag radix

Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set { 0-9, a-z } where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number radix is the radix of N1 if tag is 1, or of N2 if tag is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print Impossible. If the solution is not unique, output the smallest possible radix.

Sample Input 1:

6 110 1 10

Sample Output 1:

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

#include<iostream>
#include<cmath>
#include<algorithm>
#define MAX 10000000
using namespace std;

// 任意进制转化为十进制
long long int radix_change(string n, long long int radix){

    long long int num_10 = 0;

    for(int i = 0; i < n.length(); i++){


        if(n[i] >= '0' && n[i] <= '9'){

            num_10 = num_10 * radix + n[i] - '0';
            

        }else if(n[i] >= 'a' && n[i] <= 'z'){

            num_10 = num_10 * radix + n[i] - 'a' + 10;
            
        }

    }

    return num_10;
}

long long int find_min_base(string n){

    long long int min_base = 0;
    long long int temp;
    for(int i = 0; i < n.length(); i++){


        if(n[i] >= '0' && n[i] <= '9'){

            temp = n[i] - '0';
            

        }else if(n[i] >= 'a' && n[i] <= 'z'){

            temp = n[i] - 'a' + 10;
            
        }

        min_base = max(min_base, temp);

    }

    return min_base + 1;
}

int search_base(long long int a, string n, long long int low, long long int high){

    if(low < 2) 
        low = 2;
    long long int left = low, right = high, middle = (low + high) / 2;
    while (left <= right)
    {
       long long int b = radix_change(n, middle);
       if(b > a || b < 0){

           right = middle - 1;

       }else if(b < a)
       {
           left = middle + 1;
       }else
       {
           return middle;
       }
       middle = (left + right) / 2;
       
    }
    
    return -1;
}


int main(){

    string n1,n2;
    int tag;
    long long int radix;
    long long int a;
    long long int min_base, max_base;
 
    cin >> n1 >> n2 >> tag >> radix;
    
    if(tag == 2)
        swap(n1, n2);
    a = radix_change(n1, radix);
    min_base = find_min_base(n2);
    max_base = max(a, min_base);
    int flag = search_base(a, n2, min_base, max_base);
    if(flag == -1)
         cout << "Impossible";
    else
    {
        cout << flag;
    }
    




   

    return 0;
}