[csp-j2020] excellent splitting

[CSP-J2020] Excellent split

Title Description

Generally speaking , A positive integer can be split into the sum of several positive integers .

for example ,$1=1$,$10=1+2+3+4$ etc. . For positive integers $n$ A specific split of , Let's call it alpha “ first-class ”, If and only if under this split ,$n$ It's broken down into a number of Different Of $2$ Of Positive integer The next power . Be careful , a number $x$ Can be expressed as $2$ The positive integer power of , If and only if $x$ Can pass a positive integer $2$ Multiply together to get .

for example ,$10=8+2=2^3+2^1$ Is an excellent split . however ,$7=4+2+1=2^2+2^1+2^0$ It's not an excellent split , because $1$ No $2$ The positive integer power of .

Now? , Given a positive integer $n$, You need to judge all the splits of this number , Is there a good split . If exist , Please give us the specific split plan .

Input format

There is only one line of input , An integer $n$, Represents the number to be judged .

Output format

If all the splits of this number , There are excellent splits . that , You need to output every number in this split from large to small , Separate two adjacent numbers with a space . Can prove that , After specifying the order of splitting numbers , The splitting scheme is unique .

If there is no good split , Output -1.

Examples #1

The sample input #1

6

Sample output #1

4 2

Examples #2

The sample input #2

7

Sample output #2

-1

Tips

Examples 1 explain

$6=4+2=2^2+2^1$ Is an excellent split . Be careful ,$6=2+2+2$ Not a good split , Because it's split into $3$ The number is not satisfied, and each number is different from each other .

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Data scale and agreement

• about $20\%$ The data of ,$n\le 10$.
• For another $20\%$ The data of , Guarantee $n$ It's odd .
• For another $20\%$ The data of , Guarantee $n$ by $2$ The positive integer power of .
• about $80\%$ The data of ,$n\le 1024$.
• about $100\%$ The data of ,$1\le n\le {10}^7$.

//Author:PanDaoxi
#include <iostream>
#include <cmath>
using namespace std;
int main(){
    
    int n,a[27]={
    1};
	cin>>n;
	if(n%2==1||n==2){
    
		cout<<"-1"<<endl;
	}
	else{
    
		for(int i=1;i<=26;i++) a[i]=a[i-1]*2;
		for(int i=26;i>=1;i--){
    
			if(n>=a[i]){
    
				n-=a[i];
				cout<<a[i]<<" ";
			}
		}
	}
	return 0;
}

done .