PAT甲级——A1010 Radix

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 <iostream>
 2 #include <string>
 3 #include<math.h>
 4 #include <cctype>
 5 #include <algorithm>
 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         else
44             l = m + 1;
45     }
46     if (l > r)
47         cout << "Impossible" << endl;
48     else
49         cout << m << endl;
50     
51     return 0;
52 
53 }

来源：https://www.cnblogs.com/zzw1024/p/11178506.html

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