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 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

注意点：1.left<=right;
　　　　2.数据范围用long long



  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 
  4 long long digit[256];
  5 
  6 const long long inf = (1LL<<63)-1;
  7 
  8 
  9 
 10 void init(){
 11     
 12     for(char c='0';c<='9';c++)
 13         digit[c] = c-'0';
 14         
 15         
 16     for(char c='a';c<='z';c++)
 17         digit[c] = c-'a'+10;
 18 }
 19 
 20 long long convertNum10(string str1,long long radix,long long t){
 21     long long ans=0;
 22     
 23     long long len = str1.size();
 24     
 25     for(long long i=0;i<len;i++){
 26         char c=str1[i]; 
 27         ans=ans*radix+digit[c];
 28         
 29         if(ans<0||ans>t) return -1;        
 30     }
 31     
 32     return ans;
 33     
 34 }
 35 
 36 
 37 long long cmp(string str2,long long radix,long long t){
 38     long long n2=convertNum10(str2,radix,t);
 39     if(n2<0)return 1;
 40     else if(n2==t) return 0;
 41     else if(n2>t) return 1;
 42     else return -1;
 43     
 44 }
 45 
 46 long long binary_search(string &str2,long long low,long long high,long long t){
 47     long long left=low,right=high;
 48     long long mid;
 49     
 50     while(left<=right){
 51         mid=(left+right)/2;
 52         
 53         long long flag=cmp(str2,mid,t);
 54         
 55         if(flag<0)left =mid+1;
 56         else if(flag>0)right = mid-1;
 57         else return mid;
 58     }
 59     
 60     
 61     return -1;
 62 
 63 }
 64 
 65 
 66 long long findLargest(string &str){
 67     long long ans=-1;
 68     long long len=str.size();
 69     
 70     for(long long i=0;i<len;i++){
 71         if(digit[str[i]]>ans)
 72             ans=digit[str[i]];
 73     }
 74     
 75     
 76     return ans+1;
 77     
 78     
 79 }
 80 
 81 
 82 int main(){
 83     string str1,str2;
 84     long long tag,radix;
 85     
 86     cin>>str1>>str2>>tag>>radix;
 87     
 88     init();
 89     
 90     if(tag==2)
 91         swap(str1,str2);
 92         
 93         
 94     long long t=convertNum10(str1,radix,inf);
 95     
 96     long long low = findLargest(str2);
 97     
 98     long long high=max(low,t)+1; 
 99     
100     long long ans=binary_search(str2,low,high,t);
101     
102     
103     if(ans==-1)cout<<"Impossible\n";
104     else cout<<ans<<endl;
105     
106 }

 

posted @ 2019-07-08 20:46  鸥鹭忘机  阅读(129)  评论(0)    收藏  举报