Problem:
 Score : 200 points    Problem Statement A sequence a=\{a_1,a_2,a_3,......\} 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.    a_i = s when i = 1 , and a_i = f(a_{i-1}) when i > 1 .    Find the minimum integer m that satisfies the following condition:   There exists an integer n such that a_m = a_n (m > n) .       Constraints  1 \leq s \leq 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 .         Input Input is given from Standard Input in the following format:  s       Output Print the minimum integer m that satisfies the condition.        Sample Input 1 8      Sample Output 1 5  a=\{8,4,2,1,4,2,1,4,2,1,......\} . As a_5=a_2 , the answer is 5 .       Sample Input 2 7      Sample Output 2 18  a=\{7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,......\} .       Sample Input 3 54      Sample Output 3 114

Solution (Python):
import sys
import math
import collections
import itertools
import array
import inspect


sys.setrecursionlimit(1000000)



def chkprint(*args):
    names = {
        id(v): k
        for k, v in inspect.currentframe().f_back.f_locals.items()
    }
    print(', '.join(
        names.get(id(arg), '???') + ' = ' + repr(arg) for arg in args))



def to_bin(x):
    return bin(x)[2:]


def li_input():
    return [int(_) for _ in input().split()]


def gcd(n, m):
    if n % m == 0:
        return m
    else:
        return gcd(m, n % m)


def gcd_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = gcd(v, L[i])

    return v


def lcm(n, m):
    return (n * m) // gcd(n, m)


def lcm_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = lcm(v, L[i])

    return v



def wfs_d(D, N, K):
    """
    D: 隣接行列(距離付き)
    N: ノード数
    K: 始点ノード
    """

    dfk = [-1] * (N + 1)
    dfk[K] = 0

    cps = [(K, 0)]
    r = [False] * (N + 1)
    r[K] = True
    while len(cps) != 0:
        n_cps = []
        for cp, cd in cps:
            for i, dfcp in enumerate(D[cp]):
                if dfcp != -1 and not r[i]:
                    dfk[i] = cd + dfcp
                    n_cps.append((i, cd + dfcp))
                    r[i] = True

        cps = n_cps[:]

    return dfk



def dfs_d(v, pre, dist):
    """
    v:  現在のノード
    pre: １つ前のノード
    dist: 現在の距離

    以下は別途用意する
    D: 隣接リスト(行列ではない)
    D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
    """

    global D
    global D_dfs_d

    D_dfs_d[v] = dist

    for next_v, d in D[v]:
        if next_v != pre:
            dfs_d(next_v, v, dist + d)

    return


def sigma(N):
    ans = 0
    for i in range(1, N + 1):
        ans += i
    return ans


def comb(n, r):
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n

    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]

    for p in range(2, r + 1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p - 1, r, p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot

    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])

    return result

def bisearch(L, target):
    low = 0
    high = len(L) - 1
    
    while low <= high:
        mid = (low + high) // 2
        guess = L[mid]
        if guess == target:
            return True
        elif guess < target:
            low = mid + 1
        elif guess > target:
            high = mid - 1
    if guess != target:
        return False



dp = None

def f(k):
    if k % 2 == 0:
        return k // 2
    else:
        return 3 * k + 1

def main():
    s = int(input())

    L = [s]

    for i in range(2, 10**10):
        s = f(s)

        if s in L:
            print(i)
            return
        
        L.append(s)



main()

Solution (Python):
S = int(input())
A = set([])
while S not in A:
    A.add(S)
    if S%2==0:
        S //= 2
    else:
        S = S*3+1

print(len(A)+1)

Solution (Python):



import array
from bisect import *
from collections import *
import fractions
import heapq 
from itertools import *
import math
import random
import re
import string
import sys

s = int(input())
i = 1
As = set()
As.add(s)
prev = s
while True:
    i += 1
    if prev % 2:
        s = 3 * prev + 1
    else:
        s = prev // 2
    if s in As:
        print(i)
        break
    As.add(s)
    prev = s
8

Solution (Python):
def f(n):
    if n%2 == 0:
        return n//2
    return 3*n+1

s = int(input())
ans = 1
used = set()
while True:
    if s in used:
        break

    used.add(s)
    ans += 1
    s = f(s)
print(ans)

Solution (Python):
def f(n):
    if n%2==0:
        return n//2
    else:
        return 3*n+1

s=int(input())
ans=2
if s==1 or s==2:
    print(4)
    exit()
while(s>1):
    s=f(s)
    ans+=1
print(ans)