题目描述
A sequence a={a1,a2,a3,…} is determined as follows:
The first term s is given as input.
Let f(n) be the following function:
f(n)=n/2 if n is even, and f(n)=3n+1 if n is odd.
ai=s when i=1, and ai=f(ai−1) when i>1.
Find the minimum integer m that satisfies the following condition:
There exists an integer n such that am=an(m>n).
Constraints
·1≤s≤100
·All values in input are integers.
·It is guaranteed that all elements in a and the minimum m that satisfies the condition are at most 1000000.
输入
Input is given from Standard Input in the following format:
s
输出
Print the minimum integer m that satisfies the condition.
样例输入
8
样例输出
5
提示
a={8,4,2,1,4,2,1,4,2,1,…}. As a5=a2, the answer is 5.
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
ll num[maxn];
start{
int n=read;
int cnt = 0;
map<int,int>mp;
while(1){
cnt++;
if(mp[n])break;
mp[n]++;
if(n%2 == 0)n /= 2;
else n = 3*n+1;
}
cout<<cnt;
end;
}