pat 1010 Radix

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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 tagis 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

Problem Solving:
The min radix must bigger than the minimun num in the num that need to caculate its radix, and the max radix must no more than value of the num which has been given radix.
The way to find unknown radix is dichotomy , cause if use exhaustive method, one case will out of time. And you also need to consider if the value of num is out of range.
Here the source of this idea:https://blog.csdn.net/wyg1997/article/details/75513946

Code:

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
LL toNum(char c)
{
   if (c >= ‘0‘ && c <= ‘9‘)
      return c - ‘0‘;
   return c - ‘a‘ + 10;
}
LL toDecimal(string str, LL radix)
{
   LL num = 0;
   for (int i = 0; i < str.length(); i++)
   {
      num = num * radix + toNum(str[i]);
      if (num < 0)
         return -1;
   }
   return num;
}
int main()
{
   string a, b;
   int tag, radix;
   LL l, r, mid;
   cin >> a >> b >> tag >> radix;
   if (tag == 2)
   {
      swap(a, b);
   }
   LL base = toDecimal(a, radix);
   l = 2;
   r = base;
   for (int i = 0; i < b.size(); i++)
   {
      l = max(l, toNum(b[i] + 1));
   }
   while (r >= l)
   {
      mid = (l + r) >> 1;
      LL t = toDecimal(b, mid);
      if (t >= base || t == -1)
         r = mid - 1;
      else
         l = mid + 1;
   }
   if (toDecimal(b, l) == base)
      cout << l;
   else
      puts("Impossible");
   return 0;
}

Impression:

I really appreciate the solve of this problem, it has a really clear thinking, and functional separation is also very good. I‘ll learn from the author.

pat 1010 Radix

标签：real   unique   case   oss   osi   bsp   style   nts   href   

原文地址：https://www.cnblogs.com/stormax/p/11070630.html