一、題目

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

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2

Sample Input 1:

6 110 1 10

Sample Output 1:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

二、題目大意

給定n1、n2兩個數，求可以是兩者相等的進位制；如果沒有，輸出Impossible

三、考點

二分法、取值範圍

PS. 這道題目比較坑，沒有對取值範圍說明，int 肯定會越界，long long int 可能也會；最大取值範圍也沒有說明

四、解題思路

1、如果 tag == 2 ，交換，之後遇到類似的題目，也可以這樣處理；

2、求 n1 的10進位制數值，按照二分法求解 n2 對應 radix 下的數值，判斷是否相等，相等的話，退出，不相等的話，更新 left 和right 的值；

3、在二分法求解的過程中，long long int 可能會越界，如果求得的值 < 0 說明越界，radix 過大。這也是測試系統的一個問題，有可能全都越界，這種情況很難處理。

五、程式碼

#include<iostream>
#include<algorithm>
#include<string>
#include<math.h>
using namespace std;
long long int to10(string s, int radix) {
	long long int sum = 0;
	for (int i = 0; i < s.length(); ++i) {
		if (isdigit(s[i]))
			sum = sum*radix+s[i] - '0';
		else if (isalpha(s[i]))
			sum = sum*radix+s[i] - 'a'+10;
	}
	return sum;
}
long long int findRadix(long long int n, string s, long long int left, long long int right) {
	long long int m;
	while (left <= right) {
		long long int mid = (left + right) / 2;
		m = to10(s, mid);
		if (m == n) {
			return mid;
		}
		//進位制太大
		else if (m > n||m<0) 
			right = mid-1;
		//進位制太小
		else 
			left = mid+1;
	}
	return -1;
}
int main() {
	string s1, s2;
	//讀取資料，預處理
	int tag, radix;
	cin >> s1 >> s2 >> tag >> radix;
	if (tag == 2)
		swap(s1, s2);

	//s1轉10進位制
	long long int n1, n2;
	n1 = to10(s1, radix);

	//尋找s2的最小進位制
	long long int min_radix = 0;
	for (int i = 0; i < s2.length(); ++i) {
		if (isdigit(s2[i])) {
			if (s2[i] - '0' > min_radix)
				min_radix = s2[i] - '0'+1;
		}
		else if (isalpha(s2[i])) {
			if (s2[i] - 'a' + 10 > min_radix)
				min_radix = s2[i] - 'a' + 10+1;
		}
	}

	//二分法
	long long int max_radix = max(n1, min_radix);
	long long int ans = findRadix(n1, s2, min_radix, max_radix);

	//輸出結果
	if (ans == -1)
		cout << "Impossible";
	else
		cout << ans;
	system("pause");
	return 0;
}