1010 Radix （25 分)

2019-03-08

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:

1	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:

1	6 110 1 10

Sample Output 1:

Sample Input 2:

1 ab 1 2

Sample Output 2:

1	Impossible

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出n1,n2,tag,radix，问是否存在某一进制，使其中一个数在该进制下与tag代表的那个数在radix进制下相等。如果存在多个进制符合要求，那么输出最小的那个。

分析

进制转换问题，但是这里寻找进制必须使用二分法，否则会超时。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
#include <cmath>

using namespace std;

long long cal(string s1, long long radix){
    int l=s1.length();
    long long num=0;
    int temp;
    for(int i=0;i<l;i++){
        if(s1[i]<='9'&&s1[i]>='0')
            temp=(s1[i] - '0');

        else if(s1[i]>='a'&&s1[i]<='z') {
            temp=(s1[i] - 'a' + 10);
        }
        num += temp * pow(radix, l - i - 1);
    }
    return num;
}

long long find_radix(string s1,long long num){
    char ch=*max_element(s1.begin(),s1.end());
    long long low=(isdigit(ch)?ch-'0':ch-'a'+10)+1;
    long long high=max(num,low);
    long long t;
    while(low<=high){
        long long mid=low+(high-low)/2;
        t=cal(s1,mid);


        if(t==num)
            return mid;
        else if(t>num||t<0)
            high=mid-1;
        else
            low=mid+1;
    }
    return -1;
}

int main() {
    string s[3];
    int tag,last;
    long long radix;
    cin>>s[1]>>s[2]>>tag>>radix;
    long long  num[3]={};

    num[tag]=cal(s[tag],radix);
    int other=(tag==1?2:1);
    long long low,high;

    radix=find_radix(s[other],num[tag]);
    if(radix!=-1)
        cout<<radix;
    else
        cout<<"Impossible";
}