PAT:1010 Radix (25分)(逻辑）

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 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

柳神强！，这道题逻辑真的要清晰，每一步都得知道要干什么，首先得转换为10进制，再要找到另一个进制，另一个进制怎么着呢，首先它必然大于它所对的数的，某个位上的最大值，比如198，那么这个进制必然大于9，所以用二分法，low就确定了，那么high自然就是另一个数了，如果在另一个数的范围内还找不到，显然就是不存在了，如果假设100，如果我们在100进制内没有找到，那么下一个就是101，显然，就算这个数是1，十进制后也是101，显然不可能大于100，所以上限确定

关于二分运算，确实做得很少，关于左开右闭，与左闭右闭理解的不透彻，等我力扣刷完dp，就刷二分吧

#include<bits/stdc++.h>
using namespace std;
long long convert(string n,long long radix)
{
    long long sum=0;
    int temp=0,index=0;
    for(int i=n.length()-1;i>=0;i--){
        temp=isdigit(n[i])?n[i]-'0':n[i]-'a'+10;
        sum+=temp*pow(radix,index++);
    }
    return sum;
}
long long find_radix(string n,long long sum)
{
    char it=*max_element(n.begin(),n.end());
    long long low=isdigit(it)?it-'0':it-'a'+10;
    low++;
    long long high=max(low,sum);
    while(low<=high)
    {
        long long mid=(low+high)/2;
        long long t=convert(n,mid);
        if(t<0||t>sum) high=mid-1;
        else if(t==sum) return mid;
        else
            low=mid+1;
    }
    return -1;
}
int main()
{
    string n1,n2;
    long long tag,radix,result_radix;
    cin>>n1>>n2>>tag>>radix;
    result_radix=tag==1?find_radix(n2,convert(n1,radix)):find_radix(n1,convert(n2,radix));
    if(result_radix==-1)
        printf("Impossible");
    else
        printf("%lld",result_radix);
}