1010. Radix (25)

1010. Radix (25)

時間限制

400 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

Standard

作者

CHEN, Yue

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:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

這個問題的難點在於int會越界，就算是long也有可能越界，部分代碼參照了http://blog.csdn.net/acm_ted/article/details/20293167

//
//  main.cpp
//  Test
//
//  Created by 秦傳慶 on 16/2/25.
//  Copyright © 2016年 秦傳慶. All rights reserved.
//

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <sstream>

using namespace std;

int charToInt(char datain)
{
    int result = 0;
    if (datain >= 'a' && datain <= 'z'){
        result = datain - 'a' + 10;
    }
    else {
        result = datain - '0';
    }
    return result;
}

long anyToDecimal(string datain, long radix){
    long digit = 0;
    long radix_temp = 1;
    long number = 0;
    for (int index = static_cast<int>(datain.size()) - 1; index >= 0; index--){
        number = charToInt(datain[index]);
        digit += number * radix_temp;
        radix_temp *= radix;
        if (digit < 0){ //
            return -1;
        }
    }
    return digit;
}

int radixMayBe(string datain){
    int min_radix = 0;
    for (int i = 0; i < datain.size(); i++){
        min_radix = max(min_radix, charToInt(datain[i]));
    }
    return min_radix + 1;
}

bool binary_search(long N1, string N2, long& lower, long& upper){
    long temp;
    long middle = 0;
    bool result = false;
    while (lower <= upper){
        middle = (lower + upper) / 2;
        
        temp = anyToDecimal(N2, middle);
        if (temp > N1 || -1 == temp){
            upper = middle - 1;
        }
        else if (temp < N1){
            lower = middle + 1;
        }
        else {
            result = true;
            break;
        }
        //cout << lower << ":" << middle << ":" << upper << endl;
    }
    if (result){
        cout << middle << endl;
    } else {
        cout << "Impossible" << endl;
    }
    return result;
}

int main() {
    string N1;
    long N1_long = 0;
    string N2;
    int tag;
    long radix;
    cin >> N1;
    cin >> N2;
    cin >> tag;
    cin >> radix;
    if (2 == tag){
        swap(N1, N2);
    }
    N1_long = anyToDecimal(N1, radix);
    long min_radix = radixMayBe(N2);
    long max_radix = N1_long + 1;
    binary_search(N1_long , N2, min_radix, max_radix);
    //cout << anyToDecimal("ff", 17) << endl;
    return 0;
}