題目描述

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.

輸入描述

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.

輸出描述

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.

輸入例子1

6 110 1 10

輸出例子1

2

輸入例子2

1 ab 1 2

輸出例子2

Impossible

我的代碼

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm> 
using namespace std;
typedef long long LL;
long long Map[256]; 
long long inf=(1LL << 63)-1;

void init(){  //對'0'-'9' 'a'-'z'進(jìn)行初始化賦值 
    for(char c='0';c<='9';c++){
        Map[c]=c-'0';
    } 
    for(char c='a';c<='z';c++){
        Map[c]=c-'a'+10;   
    }
}

long long convertNum10(char a[],long long radix,long long t){  //轉(zhuǎn)換為10進(jìn)制 
    long long 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;  //判斷溢出 
    } 
    return ans;
}

long long findLargestDigit(char n2[]){  //找到n2的下界 
    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;
}

int cmp(char n2[],long long radix,long long t){  //將n2轉(zhuǎn)換為10進(jìn)制之后與t進(jìn)行判斷 
    int len=strlen(n2);
    long long num=convertNum10(n2,radix,t);
    if(num<0) return 1; 
    if(t>num) return -1;
    else if(t==num) return 0;
    else return 1;
}

long long binarySearch(char n2[],long long left,long long right,long long t){  //二分法查找進(jìn)制 
    long long mid;
    while(left<=right){
        mid=(left+right)/2;
        int flag=cmp(n2,mid,t);
        if(flag==0) return mid;
        else if(flag==-1) left=mid+1;
        else right=mid-1;
    } 
    return -1;
}


int main(){
    char n1[20],n2[20],temp[20];
    int tag,radix;
    cin>>n1;
    cin>>n2;
    cin>>tag;
    cin>>radix;
    init();  
    if(tag==2){  //將確定進(jìn)制的數(shù)放在n1 
        strcpy(temp,n1);
        strcpy(n1,n2);
        strcpy(n2,temp);
    }
    long long t=convertNum10(n1,radix,inf);  //將n1轉(zhuǎn)換為10進(jìn)制數(shù) 
    long long low=findLargestDigit(n2);  //找到n2中最大的數(shù) 
    long long high=max(t,low)+1;  
    long long ans=binarySearch(n2,low,high,t); //求解n2的進(jìn)制 
    if(ans==-1) cout<<"Impossible";
    else cout<<ans;
    return 0;
}