4.5 二分：A1010 Radix

A1010 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 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

---

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
LL Map[256];
LL INF=(1LL<<63)-1;//long long的最大值为2^63-1

void init(){
    for(char c='0';c<='9';c++){
        Map[c]=c-'0';
    }
    for(char c='a';c<='z';c++)
        Map[c]=c-'a'+10;
}
LL convertNum10(char a[],LL radix,LL t){
    LL ans=0;
    int len=strlen(a);
    for(int i=0;i<len;i++){
        ans=ans*radix+Map[a[i]];
        if(ans<0 || ans>t) return -1;//溢出或超过N1的十进制
    }
    return ans;
}
int cmp(char N2[],LL radix,LL t){
    int len=strlen(N2);
    LL num=convertNum10(N2,radix,t);
    if(num<0) return 1;//溢出，N2>t
    if(t>num) return -1;
    else if(t==num) return 0;
    else return 1;
}
int findLargestDigit(char N2[]){
    int ans=-1,len=strlen(N2);
    for(int i=0;i<len;i++){
        if(Map[N2[i]]>ans){
            ans=Map[N2[i]];
        }
    }
    return ans+1;//最大数位为ans，说明进制数的底线是ans+1
}

LL binarySearch(char N2[],LL left,LL right,LL t){
    LL mid;
    while(left<=right){
        mid=(left+right)/2;
        int flag=cmp(N2,mid,t);
        if(flag==0)return mid;//找到解，返回mid
        else if(flag==-1) left=mid+1;//往右子区间继续查找
        else right=mid-1;//往左子区间继续查找
    }
    return -1;//解不存在
}
char N1[20],N2[20],temp[20];
int tag,radix;
int main(){
    init();
    scanf("%s %s %d %d",N1,N2,&tag,&radix);
    if(tag==2){
        strcpy(temp,N1);
        strcpy(N1,N2);
        strcpy(N2,temp);
    }
    LL t=convertNum10(N1,radix,INF);
    LL low=findLargestDigit(N2);
    LL high=max(low,t)+1;
    LL ans=binarySearch(N2,low,high,t);
    if(ans==-1) printf("Impossible\n");
    else printf("%lld\n",ans);
    return 0;
}