CF-gym-100523-C(水题)

Will It Stop?

Available memory: 64 MB.

   Byteasar was wandering around the library of the University of Warsaw and at one of its facades he noticed a piece of a program with an inscription “Will it stop?”. The question seemed interesting, so Byteasar tried to tackle it after returning home. Unfortunately, when he was writing down the piece of code he made a mistake and noted:

　　　　 while n > 1 do

　　　　　　 if n mod 2 = 0 then

　　　　　　　　n := n=2

　　　　　　 else

　　　　　　　　n := 3 · n + 3

   Byteasar is now trying to figure out, for which initial values of the variable n the program he wrote down stops. We assume that the variable n has an unbounded size, i.e., it may attain arbitrarily large values.

Input

The first and only line of input contains one integer n (2<=n<=10^14).

Output

In the first and only line of output you should write a single word TAK (i.e., yes in Polish), if the program stops for the given value of n, or NIE (no in Polish) otherwise.

Example

For the input data: 4

the correct result is: TAK

 

分析：根据题目给的循环，在里面加一句，如果当前的n已经出现过，接下来又是重复了，则会一直死循环，就直接退出。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
using namespace std;
#define INF 100000
typedef unsigned long long ul;
const int maxn=10010;
map<ul,int> mp;
int main()
{
    ul n;
    scanf("%I64u",&n);
    int flag=1;
    while(n>1){
        mp[n]=1;
        if(n%2==0) n=n/2;
        else n=3*n+3;
        if(mp[n]==1) {flag=0;break;}
    }
    if(flag==1) printf("TAK\n");
    else printf("NIE\n");
    return 0;
}