PAT Advanced 1010 Radix（25）

题目描述：

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

算法描述：进制转换，二分

#include<iostream>
#include<string>
#include<ctype.h>
using namespace std;

typedef long long ll;
ll n1, n2, tag, radix;
string s1, s2;

ll trans(char c)
{
	if (isdigit(c)) return c - '0';
	else    return c - 'a' + 10;
}

ll value(string s, long long r)
{
	ll j = 0;
	for (int i = 0; i < s.size(); i ++) {
		j = j * r + trans(s[i]);
	}
	return j;
}

int main()
{
	cin >> s1 >> s2 >> tag >> radix;
	if (tag == 2) 
        swap(s1, s2);
    
	n1 = value(s1, radix);
    //将radix变为第二个数最大位+1
	radix = 0;
	for (int i = 0; i < s2.size(); i++) 
		if (radix <= trans(s2[i]))  radix = trans(s2[i]) + 1;

	for (ll step = 1; n2 = value(s2, radix);) 
    {
		if (n2 < n1) //二进制优化
        {
			radix += step;
			step *= 2;
		}
		else if (n2 == n1) 
        {
			cout << radix;
			return 0;
		}
		else 
        {
			if (step <= 2) 
            {
				cout << "Impossible";
				return 0;
			}
			else // step > 2时才存在跳过进制的情况
            {
				radix -= step / 2 - 1;
				step = 2;
			}
		}
	}
    
}