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:

2

Sample Input 2:

1 ab 1 2

Sample Output 2:

Impossible

Submit:

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
//目标：把n1的radix进制的转为10进制，再根据二分查找，尝试寻找n2对应的mid进制转为10进制比较相等则输出mid（最终所求）
//进制转换，把radix进制的s转为10进制
long long convert(string s, int radix) {//转化过程可能溢出 ，所以用long long 类型
    long long i, ans=0;//10进制结果
    for (i=0;i<s.length();i++) {
        char t = s[i];
        ans = isdigit(t) ? ans*radix+t-'0' : ans*radix + t-'a'+10;
    }
    return ans;
}
//二分查找
long long binSearch(string s, long long num) {
    char it = *max_element(s.begin(), s.end());//求出字符串中的最大字符
    long long low = (isdigit(it) ? it -'0' : it -'a' + 10)+1;//把最大字符转为数字或字母 再加1防止进位 
    long long high = max(low, num);///求最大值和已知的10进制数的最大值 如 8 8 1 10
    while (low <= high) {
        long long mid = (low+high)/2;
        long long t = convert(s,mid);//把第二个字符串以mid进制转为10进制进行比较
        if (t<0 || t>num) high = mid-1;
        else if(t == num) return mid;//相等表示是 mid进制
        else low = mid +1;//t<num 从后半部分取
    }
    return -1;
}
int main(){
    long long tag,radix,result;
    string n1,n2;
    cin >> n1 >> n2 >> tag >> radix ;
    long long temp = convert(n1,radix);
    result = tag == 1 ? binSearch(n2,convert(n1,radix)) : binSearch(n1,convert(n2,radix));//tag=1,二分查找n2,和radix进制的n1转为十进制
    if (result != -1) printf("%lld",result);
    else printf("Impossible");
    return 0;
}

参考：

柳婼-https://blog.csdn.net/liuchuo/article/details/54561626

昵称五个字-https://blog.csdn.net/a617976080/article/details/89676670