一道比较简单的递归题

#include <iostream>
#include <cmath>
using namespace std;

void out(long long x)
{
    for(int i = 60;i >= 0;i--) //最大2^60(咋可能,我随便蒙的,啥都行,别太小)
    {
        if(pow(2,i) <= x)//如果2的i次方小于x,因为是逆着枚举的,所以这里是在找当2的i次方小于x时,最大的i
		{
            if(i == 1) //如果是2的一次方不用输出2(1),直接输出2
			{
				cout << 2;
			}
			
            else if(i == 0)//如果是二的0次方也是直接输出
			{
				cout << "2(0)";
			}
			
            else//正常情况,还要继续找
			{
                cout << "2(";
	            out(i);//再继续往下搜
	            cout << ')';
            }
            
            x -= pow(2,i);//x减去这个数
            
            if(x != 0)//如果x减完后居然还有(???),简单地输出一个'+'
			{
				cout << '+';
			}
        }
    }
}

int main()
{
	long long n;//输入,要开long long,别学我还用int
    cin >> n;
    out(n);
    return 0;//返回
}

#include <queue> 
#include <math.h> 
#include <stack> 
#include <stdio.h>
#include <iostream>
#include <vector> 
#include <iomanip> 
#include <string.h> 
#include <algorithm>
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N=1e5+10;
const int INF=0x3f3f3f3f;
void f(int n)
{
	int a[30];
	int len = 0;
	while(n)
	{
		a[len++]=n%2;
		n/=2;	
	}
	int flag = 0;
	for(int i = len-1;i >= 0;i--)
	{
		if(a[i]!=0)
		{
			if(flag == 1)
				cout <<"+";
			flag=1;
			if(i==0)
				cout <<"2(0)";
			else if(i == 1)
				cout << "2";
			else if(i == 2)
				cout << "2(2)";
			else
			{
				cout << "2(";
				f(i);
				cout << ")";
			} 
		}
		
	}
}
int main()
{
	int n;
	cin>>n;
	f(n);
	return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <math.h>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <stack>

using namespace std;

#define LL long long
const int N =1e5+10;
const int INF =0x3f3f3f3f;
void f(int n){
	int a[40];
	int len=0;
	while(n){
		a[len++]=n%2;
		n/=2;
	}
	for(int i=len-1,flag=0;i>=0;i--){
		if(a[i]!=0)
		{
			if(flag)
				cout<<"+";
			flag=1;
			if(i==0)
				cout<<"2(0)";
			else if(i==1)
				cout<<"2";
			else if(i==2)
				cout<<"2(2)";
			else{
				cout<<"2(";
				f(i);
				cout<<")";
			}
		}
	}	
}
int main(){
	int n;
	cin>>n;
	f(n);
return 0;
}