不想思考，直接打表：

#include<iostream>
#include<vector>
typedef long long LL;
#define _P(A) A=-~A
#define fione(i,a,b) for(int i=a;i<=b;_P(i))
#define Fione(i,a,b) for(int i=a;i<b;_P(i))
int n;
std::vector<int>a;
inline void work(int id){
	switch(id){
		case 0:std::cout<<"2(0)";break;
		case 1:std::cout<<"2";break;
		case 2:std::cout<<"2(2)";break;
		case 3:std::cout<<"2(2+2(0))";break;
		case 4:std::cout<<"2(2(2))";break;
		case 5:std::cout<<"2(2(2)+2(0))";break;
		case 6:std::cout<<"2(2(2)+2)";break;
		case 7:std::cout<<"2(2(2)+2+2(0))";break;
		case 8:std::cout<<"2(2(2+2(0)))";break;
		case 9:std::cout<<"2(2(2+2(0))+2(0))";break;
		case 10:std::cout<<"2(2(2+2(0))+2)";break;
		case 11:std::cout<<"2(2(2+2(0))+2+2(0))";break;
		case 12:std::cout<<"2(2(2+2(0))+2(2))";break;
		case 13:std::cout<<"2(2(2+2(0))+2(2)+2(0))";break;
		case 14:std::cout<<"2(2(2+2(0))+2(2)+2)";break;
	}
}
inline void solve(){
	bool pd(false);
	for(int i=a.size()-1;i>=0;--i){
		if(a[i]){
			if(pd)std::cout<<'+';
			pd=true;			
			work(i);
		}
	}
}
signed main(){
	std::cin.tie(nullptr)->std::ios::sync_with_stdio(false);
	std::cin>>n;
	for(;n;n>>=1)a.push_back(n&1);
	solve();
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
string run(int x,int i=0,string s=string("")){
	if(x==0)return string("0");
	do if(x&1)s=(i==1?"2":"2("+run(i)+")")+(s==""?"":"+")+s;//拼接字符串，应题意，要把低次方接在后面
	while(++i,x>>=1);//每次向右移位
	return s;
}
int main(){
	int x;cin>>x;
	cout<<run(x)<<endl;
}