Educational Codeforces Round 33 (Rated for Div. 2) B题. Beautiful Divisors

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

题意：100000 以内 有几个 美丽的数字 ， 分别是 ：

　　1(2)->1(10) 110(2)->6(10) 11100(2)->28(10) 1111000(2) -> 120(10) 111110000(2)->496(10) 11111100000(2) -> 2016(10)

1111111000000(2) -> 8128(10) 111111110000000(2) -> 32640(10) 11111111100000000(2) -> 130816(10)

注意看题 ， 题目要的是 ：“满足 是 n 的 除数 同时 是美丽数字 两个条件的最大数字” ，

/* 快速幂判断美丽数字  */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std ;

#define LL long long
// 快速幂
LL pow_x(int k ) {
LL result = 1 ;
LL a = 2 ;

while(k) {
if(k&1) {
result = result * a ;
}
a = a*a ;
k=k/2 ;
}
return result ;
}

int main() {

int n ;
while(~scanf("%d" , &n)) {
bool flag = false ;
int result ;
for(int i=n ; i>0 ; i--) {
for(int k = 1 ; ; k++ ) {
if((pow_x(k)-1)*pow_x(k-1) == i ) {
if(n%i==0) {
result = i ;
flag = true ;
break ;
}

}

if((pow_x(k)-1)*pow_x(k-1) > i ) {
break ;
}
}
if(flag) {
break;
}
}
printf("%d\n" , result) ;
}
return 0 ;

}

View Code