正经（？）题解

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首先，所谓有没有“优秀的拆分”，其实就是奇数没有优秀的拆分，偶数有！！！

你想想，奇数拆出来，必定有$2^0$(1)，是不是？？？
所以，这就是个阅读题！
判断代码：

if(n&1){//判断奇数，与n%2效果一样，但比n%2快，推荐使用。
    cout<<-1;
    return 0;
}

else……

    else{
        string s;
        while(n!=0){
            s+=(n%2)+'0';
            n/=2;
        }
        for(int i=s.size()-1;i>=0;i--){
            if(s[i]=='1'){
                cout<<int(pow(2,i))<<' ';
            }
        }
    }

先转二进制，然后就容易拆了！！！

注意点：

用pow会默认大数用e记法，得强转int。

然后就是完整代码了！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！！

#include<bits/stdc++.h>//万恶之源万能头
using namespace std;
int main(){
    ios::sync_with_stdio(false);//关闭流同步（增加效率）
    cin.tie(nullptr);//解绑cin
    cout.tie(nullptr);//解绑cout
    int n;
    cin>>n;
    if(n&1){
        cout<<-1;
        return 0;
    }
    else{
        string s;
        while(n!=0){
            s+=(n%2)+'0';
            n/=2;
        }
        for(int i=s.size()-1;i>=0;i--){
            if(s[i]=='1'){
                cout<<int(pow(2,i))<<' ';
            }
        }
    }
}

时间复杂度$O(\log_2^n+1)->O(\log_2^n)->O(\log^n)$