Arc135 a (time complexity analysis)

subject
The question : take x decomposition , Make the product of all numbers after decomposition reach the maximum . about x, It can be broken down into x/2( Round up ) and x/2( Round down )
Ideas : Memory search . When x<=4, Don't divide . otherwise , Keep dividing .
Time complexity : O(logX) or O(logX * loglogX)
  use unordered_map
hypothesis n = 2 ^ k,T(n) = T(n/2)+1.T(n) = k = logn
hypothesis n != 2^k, 2^k-1 <= n < 2^k, It can be proved that at most 2k Time f(x),T(n) < 2 * (log(x)+1) = O(log(x)).
  If the map
  It takes time to memorize logn Time for ,n At most 2(log(x)+1).logn As mentioned above loglogX Time .
Code :

// Problem: A - Floor, Ceil - Decomposition
// Contest: AtCoder - AtCoder Regular Contest 135
// URL: https://atcoder.jp/contests/arc135/tasks/arc135_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round()  rounding  ceil()  Rounding up  floor()  Rounding down 
// lower_bound(a.begin(),a.end(),tmp,greater<ll>())  First less than or equal to 
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 998244353;
const double eps = 1e-6;
inline int lowbit(int x){
     return x&(-x);}
template<typename T>void write(T x)
{
    
    if(x<0)
    {
    
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
    
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){
    if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){
    x = x*10+ch-48;ch=getchar();}x*=f;
}
#define int long long
int n,m,k,T;
map<int,int> mp;
int fun(int x)
{
    
	if(x<=4)
	{
    
	   return x;
	}
	if(mp[x]) return mp[x];
	if(x&1) return mp[x] = fun(x/2) * fun((x+1)/2) % mod;
	else return mp[x] = fun(x/2) * fun(x/2) % mod;
}
void solve()
{
    
   read(n);
   write(fun(n)); puts("");
}
signed main(void)
{
      
   T = 1;
   // OldTomato; cin>>T;
   // read(T); 
   while(T--)
   {
    
   	 solve();
   }
   return 0;
}