PAT 1010 Radix (25)

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博客围绕1010 Radix问题展开，给定一对正整数N1和N2，需在已知一个数进制的情况下求另一个数的进制。解题采用二分查找，要确定左右界，n2中最大字符对应值+1为下界，上界不超过n1进制，代码中使用unsigned long long类型。

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1010 Radix (25)（25 分）

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:

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

Sample Input 3:

12 c 1 10

Sample Output 3:

思路：

这真是我做到现在觉得最坑的一题……虽然写了是{0~9,a~z}，但是进制远远不止36！因为数字会很大，所以这里用二分查找来做，于是就要确定左右两端的值。n2中最大的字符对应的值+1是查找的下界，需要注意的是如果字符全是‘0’的话下界为2，因为最小是二进制。至于上界的话，n2的进制肯定不会超过n1进制（是n1进制不是n1的进制）。一直用long long也WA了好多遍，最后翻了别人的才知道要unsigned long long。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <map>  
#include <set>
using namespace std;

int ctoi[127];

unsigned long long convert(string s, int r)
{
	long long n = 0;
	for (int i = 0; i <s.length(); i++)
		n = n*r+ctoi[(s[i])];
	return n;
}

int find_left(string s)
{
	char max = 0;
	for (int i = 0; i < s.length(); i++)
	{
		if (s[i] > max)
			max = s[i];
	}
	if (max == '0')
		return 2;
	else
		return ctoi[max] + 1;
}

int main()
{
	for (int i = '0'; i <= '9'; i++)
		ctoi[i] = i - '0';
	for (int i = 'a'; i <= 'z'; i++)
		ctoi[i] = i - 'a' + 10;
	string s1, s2;
	int tag, radix;
	unsigned long long n1, n2, l, r, m;
	cin >> s1 >> s2 >> tag >> radix;
	if (tag == 2)
		swap(s1, s2);
	n1 = convert(s1, radix);
	l = find_left(s2);
	r = n1;
	while (l < r)
	{
		m = (l + r) / 2;
		n2 = convert(s2, m);
		if (n2 == n1)
			l = r = m;
		else if (n2 > n1)
			r = m - 1;
		else
			l = m + 1;
	}
	if (convert(s2, l) == n1)
		cout << l << endl;
	else
		cout << "Impossible" << endl;
	return 0;
}