1010 Radix

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 $N_1$ and $N_2$ 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 $N_1$ if tag is $1$, or of $N_2$ if tag is $2$.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation $N_1=N_2$ 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

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题解：

前 $10$ 题里最难的一题了，考点主要有：任意进制到十进制数的转换、二分法的应用、大数溢出，其中大数溢出比较坑。。。
本题若不使用二分法，而采用从radix = 2开始遍历的方式，则无法通过测试点7；但若使用二分法而没有判断大数溢出的情况，则会有很多测试点不通过。

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注意事项：

1. 将任意进制的数转换为十进制数：

	int sum = 0;	// 此处需要注意判断 sum 的范围, 必要时应使用 long long
	for(int i=0;num[i]!='\0';i++
		sum = sum*radix + num[i] - '0';

2. 关于大数溢出的判断方法：

	long long num;	
	......		// 此处保证正常情况下 num >= 0
	if(num < 0)
		// 溢出

3. 二分查找

	while(low<=high){
		mid = low + (high-low)/2;	// 避免 (low+high) 发生溢出
		if(...)
			low = mid + 1;
		else
			high = mid - 1;

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#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;

long long trans2dec(char* num, long long radix)
{
    long long sum = 0;
    for(int i=0;num[i]!='\0';i++){
        if(num[i]>='0' && num[i]<='9')
            sum = sum*radix + num[i] - '0';
        else
            sum = sum*radix + num[i] - 'a' + 10;
    }
    return sum;
}

int main()
{
    char num[2][12] = {'\0'};
    int tag;
    long long radix, N1,N2;
    cin>>num[0]>>num[1]>>tag>>radix;
    N1 = trans2dec(num[tag-1],radix);

    radix = 1;
    for(int i=0;num[2-tag][i]!='\0';i++){
        if(num[2-tag][i]>='0' && num[2-tag][i]<='9'){
            if(radix < num[2-tag][i] - '0')
                radix = num[2-tag][i] - '0';
        }
        else
            if(radix < num[2-tag][i] - 'a' + 10)
                radix = num[2-tag][i] - 'a' + 10;
    }

    long long low = radix + 1;
    long long high = N1 + 1;
    while(low <= high){
        radix = low + (high - low)/2;
        N2 = trans2dec(num[2-tag], radix);
        if(N1 == N2){
            cout<<radix;
            return 0;
        }
        else if(N2 < 0 || N2 > N1){     // 当 N2 溢出时，其值为负！
            high = radix - 1;
        }
        else{
            low = radix + 1;
        }
    }
    cout<<"Impossible";
}