PAT Advanced 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

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

给出N1和N2，并指定其中一者的进制，求一个使两数相等的进制。这里的进制数可能是0~N的任何一个数，并且由于进制越小，同一个字符串表示的数越小，该题可以采用二分查找的方式寻找答案。

//#include <bits/stdc++.h>
//
//using namespace std;
//
//long long convert(string &raw,int radix)
//{
//    int len=raw.length();
//    long long ans=0,p=1;
//    for(int i=len-1;i>=0;i--)
//    {
//        int r=0;
//        if(raw[i]>='0'&&raw[i]<='9')
//            r=raw[i]-'0';
//        else if(raw[i]>='a'&&raw[i]<='z')
//            r=raw[i]-'a'+10;
//        if(r>=radix)return -1;
//        ans+=r*p;
//        p*=radix;
//    }
//    return ans;
//}
//int main()
//{
//    string N1,N2;
//    int tag,radix;
//    cin>>N1>>N2>>tag>>radix;
//    if(tag==1)
//    {
//        long long  tmp=convert(N1,radix);
//        int lradix=1,rradix=max((long long)32,tmp);
//        while(lradix<rradix-1)
//        {
//            int t2radix=(lradix+rradix)/2;
//            long long t2=convert(N2,t2radix);
//            if(t2<tmp)
//            {
//                lradix=t2radix;
//            }else if(t2>=tmp)
//            {
//                rradix=t2radix-1;
//            }
//        }
//        if(convert(N2,(lradix+rradix)/2+1)==tmp){cout<<(lradix+rradix)/2+1<<endl;return 0;}
//        cout<<"Impossible"<<endl;
//    }else if(tag==2)
//    {
//        long long  tmp=convert(N2,radix);
//        int lradix=1,rradix=max((long long)32,tmp);
//        while(lradix<rradix-1)
//        {
//            int t2radix=(lradix+rradix)/2;
//            long long t2=convert(N1,t2radix);
//            if(t2<tmp)
//            {
//                lradix=t2radix;
//            }else if(t2>=tmp)
//            {
//                rradix=t2radix-1;
//            }
//        }
//                if(convert(N1,(lradix+rradix)/2+1)==tmp){cout<<(lradix+rradix)/2+1<<endl;return 0;}
//        cout<<"Impossible"<<endl;
//    }
//    return 0;
//}
#include <bits/stdc++.h>

using namespace std;

bool validate(long long radix,string &raw,long long  cmp)
{
    int len=raw.length();
    long long  ans=0,p=1;
    for(int i=len-1;i>=0;i--)
    {
        long long r=0;
        if(raw[i]>='0'&&raw[i]<='9')
            r=raw[i]-'0';
        else if(raw[i]>='a'&&raw[i]<='z')
            r=raw[i]-'a'+10;
        if(r>=radix)return false;
        ans+=r*p;
        p*=radix;
    }
    return ans==cmp;
}
long long convert(string &raw,long long radix)
{
    int len=raw.length();
    long long ans=0,p=1;
    for(int i=len-1;i>=0;i--)
    {
        long long  r=0;
        if(raw[i]>='0'&&raw[i]<='9')
            r=raw[i]-'0';
        else if(raw[i]>='a'&&raw[i]<='z')
            r=raw[i]-'a'+10;
        //if(r>=radix)return -1;
        ans+=r*p;
        p*=radix;
    }
    return ans;
}
int main()
{
    string N1,N2;
    int tag,radix;
    cin>>N1>>N2>>tag>>radix;
    string known=tag==1?N1:N2;
    string unknown=tag==1?N2:N1;
    long long tmp=convert(known,radix);
    int len=unknown.length();
    long long digitmax=0;
    for(int i=0;i<len;i++)
    {
        digitmax=max(digitmax,1+(long long)((unknown[i]>='0'&&unknown[i]<='9')?unknown[i]-'0':unknown[i]-'a'+10));
    }
    long long left=digitmax;
    long long right=max(digitmax,tmp);

    while(left<=right)
    {
        long long mid=(left+right)/2;
        long long t2=convert(unknown,mid);
        if(t2==tmp)
        {
            cout<<mid<<endl;
            return 0;
        }else if(t2<0||t2>tmp)
        {
            right=mid-1;
        }else{
            left=mid+1;
        }
    }
    cout<<"Impossible"<<endl;

    return 0;
}