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

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本文介绍了一个有趣的数学问题：如何找到一个整数的最大“美丽”除数。美丽数是指其二进制表示由连续的 k+1 个 1 后跟 k 个 0 组成的整数。文章提供了一段 C++ 代码示例，用于解决这个问题。

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 留下了不细心看题代码能力垃圾啊的眼泪刚开始没看清题写的很麻烦之后没又加等号一直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){  //这里忘记加了 
         x=b[j];
      }
      else if(b[j]>n) break;
   }
   cout<<x<<endl;
   return 0;
}