2 seconds
256 megabytes
standard input
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!
The only line of input contains one number n (1 ≤ n ≤ 105) — the number Luba has got.
Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.
3
1
992
496
【题意】：beautiful数定义：如果该数在二进制下表示由k+1个连续的1个，接着k个连续的0组成，则该数字称为beautiful的。比如：
换句话说，当且仅当存在一些正整数k使得这个 number = ( 2k - 1 ) * ( 2^(k - 1) ) ,这个number就是beautiful的。现在给你一个number，求它最大beautiful因子。
【分析】：打表得到K为最大9。求最大因子数可以逆向枚举。
【代码】：beautiful numbers表：
void init(){
num[1] = 1 ;
num[2] = 6 ;
num[3] = 28 ;
num[4] = 120 ;
num[5] = 496 ;
num[6] = 2016 ;
num[7] = 8128 ;
num[8] = 32640 ;
num[9] = 130816 ;
}
#include <bits/stdc++.h> using namespace std;
const int N = ;
int n;
int a[N];
int main()
{
for(int i=;i<=;i++) a[i]=(pow(,i)-)*pow(,i-);
scanf("%d",&n);
for(int i=;i>=;i--) if(n%a[i]==) {printf("%d\n",a[i]);break;}
return ;
}