PAT.Radix(史上最 NT 题， (其实是我脑瘫/ 哭唧唧))

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

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

/*
    本题大意：给出两个任意进制的数字，让你判断将他们都转换为十进制后是否相等。
    注意其中一个数字和他的进制已经给出，另一个数只给出数某种进制下的编码，让
    你求出这个数的最小进制数，使得这两个数字在十进制下相等。
    
    思路：题目给出2 ~ 36 其实是一个坑，答案进制数有可能超大，但其范围在
    (这个数中最大的数字 + 1，已知数字 + 1)这个范围内，知道范围，求最小值
    我们直接二分答案就行了。
*/
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#define mid ((l + r) >> 1)
using namespace std;

typedef long long ll;

ll inf = 8e18;

string s1, s2;
ll n1, n2;

ll quickly_pow(ll a, int b) {
    ll ans = 1;
    while(b) {
        if(b & 1) {
            ans *= a;
        }
        a *= a;
        b >>= 1;
    }
    return ans;
}

ll get_decimal(string s, ll radix) {
    ll ans = 0;
    int len = s.size();
    for(int i = len - 1; i >= 0; i --) {
        if(s[i] > '9') ans += (s[i] - 'a' + 10) * quickly_pow(radix, len - i - 1);
        else ans += (s[i] - '0') * quickly_pow(radix, len - i - 1);
        if(ans > n1 || ans < 0) return n1 + 1;
    }
    return ans;
}

int main() {
    int tag, radix;
    while(cin >> s1 >> s2 >> tag >> radix) {
        n1 = inf;
        if(tag == 2) swap(s1, s2);
        n1 = get_decimal(s1, radix);
        char c = *max_element(s2.begin(), s2.end());
        ll l = c < '9' ? (c - '0' + 1) : (c - 'a' + 11), r = n1 + 1, ans = inf, temp;
        while(l <= r) {
            temp = get_decimal(s2, mid);
            if(temp > n1) r = mid - 1;
            else if(temp < n1) l = mid + 1;
            else {
                ans = min(ans, mid);
                r = mid - 1;
            }
        }
        if(ans == inf) cout << "Impossible" << endl;
        else cout << ans << endl;
    }
    return 0;
}