PAT 1010 Radix (二分)

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.

思路

输出可以使得N1=N2的最小进制。

因为此题的进制上限是正无穷，因此考虑使用二分法来挑选最佳进制。因此，问题可以转化为如何选择进制的最大值与最小值。

小技巧

​ ①如果flag=2，那么swap(a,b )

​ ②用string来存字符串，统一将字符转换为对应的整数，方便运算，见代码中的init()函数（如果用char[]来存，就不能这么操作了，因为'0'-'0' = '\0'，strlen函数会出问题）。

这个题有比较坑的地方

​ ①long long类型有可能溢出。这就造成了如果N1的范围大于了long long的范围，有可能会编程负数。我看一些博客中，简单的加了一个判断条件N2 < 0。但是面对一个超范围的数字，应该不能简单的判断N1与N2的大小（例如N2 = -1，但有可能是LLONG_MAX + LLONG_MAX + 1；而N1 = 8，但有可能是LLONG_MAX + LLONG_MAX + 9，这个时候通过N2<0，来作为N2>N1的判断条件是不对的）。好在这题的数据没有考虑的这个问题。

​ ②一定要注意进制的最大值是无穷，而不是35。但是最大值是可以确定的。设N1对应的十进制为num，那么N2的最大进制不超过num。因为N2越小，最佳进制就要越大。N2的最小值为1（最佳进制下；0的话num一定也满足），那么num就已经足够了。另外要保证最大进制大于最小进制，因此代码中有个max操作。

代码

#include <stdio.h>
#include <string>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <map>
#include <limits.h>
using namespace std;
string a, b;
long long flag, radex;
long long minn = INT_MIN, maxx = INT_MIN;

long long to10(string s, long long radex){
	long long p = 1;
	long long sum = 0;
	for(int i = s.length() - 1; i >= 0; i--){
		sum = sum + p * s[i];
		p *= radex;
	}
	return sum;
}

void prove(long long dec){
	while(minn <= maxx){
		long long mid = (minn + maxx) >> 1;
		long long num = to10(b, mid);
		if(num < 0 || num > dec){
			maxx = mid - 1;
		}
		else if(num == dec){
			cout << mid;
			return;
		}
		else{
			minn = mid + 1;
		}
	}
	cout << "Impossible";
}

void init(){
	for(int i = 0; i < a.length(); i++){
		if(isdigit(a[i]))	a[i] = a[i] - '0';
		else 	a[i] = a[i] - 'a' + 10;
	}
	for(int i = 0; i < b.length(); i++){
		if(isdigit(b[i]))	b[i] = b[i] - '0';
		else 	b[i] = b[i] - 'a' + 10;
	}
}

int main() {
	cin >> a >> b >> flag >> radex;
	if(flag == 2)		swap(a, b);
	init();
	minn = *max_element(b.begin(), b.begin()) + 1;
	long long dec = to10(a, radex);
	maxx = max(dec, minn);
	prove(dec);

	return 0;
}