【PAT (Advanced Level) Practice】1010 Radix (25 分)

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 $N_1$ and $N_2$ , 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:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

---

#include<iostream>

using namespace std;

typedef long long LL;

string n1, n2;
int tag, radix;

int get(char c){
    
    if('0' <= c && c <= '9') return c - '0';
    return c - 'a' + 10;
}

LL calc(string s, LL r){
    
    LL res = 0;
    // 秦九韶算法
    for(auto c: s){
        if((double)res * r + get(c) > 1e16) return 1e18;  // 当转换成10进制超出最大范围时，这个进制一定偏大了！
        res = res * r + get(c);
    }
    
    return res;
}

int main(){
    
    cin >> n1 >> n2 >> tag >> radix;
    
    if(tag == 2) swap(n1, n2);
    
    LL target = calc(n1, radix);
    
    LL l = 0, r = target;
    for(auto c: n2) l = max(l, (LL)get(c) + 1);
    
    while(l < r){
        LL mid = l + r >> 1;
        if(calc(n2, mid) >= target) r = mid;
        else l = mid + 1;
    }
    
    if(target != calc(n2, l)) puts("Impossible"); 
    else cout << l << endl;
    
    return 0;
}