codeforces Educational Codeforces Round 33 (Rated for Div. 2)B

ACM_e發表於2017-11-24

B. Beautiful Divisors
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists of k + 1 consecutive ones, and then k consecutive zeroes.

Some examples of beautiful numbers:

  • 12 (110);
  • 1102 (610);
  • 11110002 (12010);
  • 1111100002 (49610).

More formally, the number is beautiful iff there exists some positive integer k such that the number is equal to (2k - 1) * (2k - 1).

Luba has got an integer number n, and she wants to find its greatest beautiful divisor. Help her to find it!

Input

The only line of input contains one number n (1 ≤ n ≤ 105) — the number Luba has got.

Output

Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.

Examples
input
3
output
1
input
992
output
496

mad  留下了 不細心看題 程式碼能力垃圾啊的眼淚    剛開始沒看清題 寫的很麻煩 之後沒又加 等號 一直wa  


#include<bits/stdc++.h>
using namespace std;
long long xxx(string x){
   int a[10001];
   for(int j=0;j<x.size();j++){
     a[j]=x[j]-'0';
   }
   long long l=0;
   long long i=x.size();
   for(int j=0;j<x.size();j++){
     l+=pow(2,i-j-1)*a[j];
   }
   return l;
}
int main(){
   int n;
     cin>>n;
   string a;
   int i=0;
   int b[10001];
   for(int j=1;;j++){
      for(int k=0;k<j;k++){
         a=a+"1";
      }
      for(int k=0;k<j-1;k++){
         a=a+"0";
      }
      b[j]=xxx(a);
      if(b[j]>10000000){
        i=j;
        break;
      }
      a="";
   }
   int x=0;
   for(int j=1;j<=i;j++){
      if(b[j]<=n&&n%b[j]==0){  //這裡忘記加了 
         x=b[j];
      }
      else if(b[j]>n) break;
   }
   cout<<x<<endl;
   return 0;
}



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