非正经题解

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
ll mi[30] = {0,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,
			65536,131072,262144,524288,1048576,2097152,4194304,8388608,
			16777216};//储存2的幂(可以写成函数自动定义别学我手搓2的幂())
vector<ll> ans;
ll bs(ll n){//二分遍历mi数组
	ll ans = 0;
	int l = 1,r = 24;
	while(l < r){
		int mid = (l + r) / 2;
		if(mi[mid] <= n){
			ans = mid;
			l = mid + 1;
		}else r = mid - 1;
	}
	return ans;
}
int main(){
	cin >> n;
	if(n % 2) cout << -1;//是奇数则输出-1
	else{
		int a = bs(n);
		for(int i = a + 1;i >= 1;i--){
			if(mi[i] <= n){
				ans.push_back(mi[i]);//把mi数组中小于n的最大的数字存储进ans
				n -= mi[i];//n减去上面那个数字
			}
		}
		for(int i = 0;i < ans.size();i++) cout << ans[i] << " ";//循环遍历输出
	}
	return 0;
//灵感来源于进制转换的一种方法,集训营考试的时候突发奇想写了这个东西交上去然后AC了
}
//Q=cm∆t