PAT (Advanced Level) Practise 1010 Radix (25)

1010. Radix (25)

時間限制

400 ms

記憶體限制

65536 kB

代碼長度限制

16000 B

判題程式

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

此題的坑點主要在于沒有說要你求的進制要小于36，是以數字可能很大，要用ull，雖然感覺可能還是要炸，然而資料并沒那麼強。

#include<cstdio>
#include<stack>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
typedef unsigned long long uLL;
const int maxn = 1e5 + 10;
string a, b;
int tag, radix, res;
uLL ans = 0, l, r;

uLL get(char ch)
{
  if ('0' <= ch&&ch <= '9') return ch - '0';
  return ch - 'a' + 10;
}

int main()
{
  cin >> a >> b >> tag >> radix;
  if (tag == 2) swap(a, b);
  for (int i = 0; a[i]; i++) ans = ans*radix + get(a[i]);
  for (int i = 0; b[i]; i++) l = max(l, get(b[i]));
  for (l++, r = ans + 1; l <= r;)
  {
    uLL mid = l + r >> 1;
    uLL check = 0;
    for (int i = 0; b[i]; i++) check = check*mid + get(b[i]);
    if (check == ans) res = mid;
    if (check >= ans) r = mid - 1; else l = mid + 1;
  }
  res ? printf("%d\n", res) : printf("Impossible\n");
  return 0;
}