1010 Radix

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本文介绍了一个基数转换的问题，即给定两个正整数，判断是否存在一种基数使得这两个数相等。文章详细阐述了如何通过解析输入的数值并进行基数转换运算来解决这个问题。如果存在可能的基数，则输出该基数；若不存在，则输出‘Impossible’。

1010 Radix （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

#include <bits/stdc++.h>
using namespace std;
int main()
{
	string n1,n2;
	int tag,radix;
	long long sum1=0,sum2=0;
	vector <int> v[3];
	cin>>n1>>n2>>tag>>radix;
	for(int i=0;i<n1.length();i++)
	{
		if(isdigit(n1[i]))
			v[1].push_back(n1[i]-'0');
		else
			v[1].push_back(n1[i]-87);
	}
	for(int i=0;i<n2.length();i++)
	{
		if(isdigit(n2[i]))
			v[2].push_back(n2[i]-'0');
		else
			v[2].push_back(n2[i]-87);
	}
	int flag=1;
	int k;
	if(tag==1)
		k=2;
	else
		k=1;
	for(int i=v[tag].size()-1;i>=0;i--)
	{
		sum1+=v[tag][i]*pow(radix,v[tag].size()-1-i);
	}
	long long low=*max_element(v[k].begin(),v[k].end())+1;
	long long high=max(low,sum1);
	while(low<=high)
	{
		sum2=0;
		long long mid=(low+high)/2;
		for(int j=v[k].size()-1;j>=0;j--)
		{
			sum2+=v[k][j]*pow(mid,v[k].size()-1-j);
		}
		if(sum2<0||sum2>sum1)
			high=mid-1;
		else if(sum2<sum1)
			low=mid+1;
		else
		{
			flag=0;
			cout<<mid;
			break;
		}
	}
	if(flag==1)
		cout<<"Impossible";
	return 0;
}