PAT 1010 Radix（X）

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<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 500

vector<char> vec;

map<char,int> mp;

ll tansss(string s,int x){
    ll sum = ;
    ll t = ;
    for(int i=s.size()-;i>=;i--){
        sum += mp[s[i]]*t;
        t *= x;
    }
    return sum;
}

int main(){
    for(int i=;i <= ;i++){
        vec.push_back(char(''+i));
    }
    for(int i=;i <= ;i++){
        char temp = char('a'+(i-));
        vec.push_back(temp);
    }

    for(int i=;i < vec.size();i++){
        mp[vec[i]] = i;
    }

//    while(1){
//        string s; cin >> s;
//        int radix; cin >> radix;
//        cout << tansss(s,radix);
//    }

    string s1,s2;
    int tag,radix;
    cin >> s1 >> s2 >> tag >> radix;

    if(tag == ){
        ll ok = tansss(s1,radix);
        for(int i=;i<=;i++){
            if(tansss(s2,i)==ok){
                cout << i;
                return ;
            }
        }
        cout << "Impossible";
    }
    else{
        ll ok = tansss(s2,radix);
        for(int i=;i<=;i++){
            if(tansss(s1,i)==ok){
                cout << i;
                return ;
            }
        }
        cout << "Impossible";
    }

    return ;
}

还要判断溢出。。。打扰了拿16分溜了