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 这题不难，主要是要考虑数据溢出的问题，和使用二分法去寻找匹配的二进制，而二分法中的左右边界是要考虑怎么得到的，这是个难点

1 #include 
     
       2 #include 
      
        3 #include
       
         4 #include 
        
          5 #include 
         
           6 using namespace std; 7  8 //最简单的一种一种格式试 9 int main()10 {11     string N1, N2;12     long long int tag,radix;//防止溢出13     cin >> N1 >> N2 >> tag >> radix;14     if (tag == 2)15     {16         string N = N2;17         N2 = N1;18         N1 = N;19     }20     long long int aim = 0, n = 0;//防止化为10制止时溢出21     for (int i = N1.length() - 1, j = 0; i >= 0; ++j, --i)//将N1划为10进制22     {23         n = isdigit(N1[i]) ? (N1[i] - '0') : (N1[i] - 'a' + 10);24         aim += n * pow(radix, j);25     }26     long long int l, r, m;27     char it = *max_element(N2.begin(), N2.end());//找出最大的元素28     l = (isdigit(it) ? it - '0' : it - 'a' + 10) + 1;//找到N2形成的最小进制    29     r = max(aim, l);30     while (l <= r)31     {32         m = l + (r - l) / 2;33         long long int res = 0, n;34         for (int i = N2.length() - 1, j = 0; i >= 0; ++j, --i)35         {36             n = isdigit(N2[i]) ? (N2[i] - '0') : (N2[i] - 'a' + 10);    37             res += n * pow(m, j);38         }39         if (res == aim)40             break;41         else if (res<0 || res>aim)42             r = m - 1;43         else44             l = m + 1;45     }46     if (l > r)47         cout << "Impossible" << endl;48     else49         cout << m << endl;50     51     return 0;52 53 }