HDU 5752

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 423    Accepted Submission(s): 188

Problem Description

Let's define the function $f(n)=lfloor sqrt{n} floor$.

Bo wanted to know the minimum number $y$ which satisfies $f^y(n)=1$.

note:$f^1(n)=f(n),f^y(n)=f(f^{y-1}(n))$

It is a pity that Bo can only use 1 unit of time to calculate this function each time.

And Bo is impatient, he cannot stand waiting for longer than 5 units of time.

So Bo wants to know if he can solve this problem in 5 units of time.

 

Input

This problem has multi test cases(no more than $120$).

Each test case contains a non-negative integer $n(n<10^{100})$.

 

Output

For each test case print a integer - the answer $y$ or a string "TAT" - Bo can't solve this problem.

 

Sample Input

233

233333333333333333333333333333333333333333333333333333333

 

Sample Output

3

TAT

 

Source

2016 Multi-University Training Contest 3

 

Recommend

wange2014

 题意：给你一个数 问你需要不断开根多少次结果等于1 超过5次输出 TAT

 

 题解：由于有5次的这个限制，所以尝试寻找分界点。

很容易发现是2^{32}，所以我们先比较输入的数字是否比这个大，然后再暴力开根

 

/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^    ^ ^
 O      O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
#define mod 1000000007
#define PI acos(-1.0)
using namespace std;
char a[205];
ll ans;
int main()
{
      while(scanf("%s",a)!=EOF){
        ans=0;
        int len=strlen(a);
        if(len>10||(len==1&&a[0]=='0'))
         printf("TAT
");
        else
        {
            for(int i=0;i<len;i++)
                ans=ans*10+a[i]-'0';
            int jishu=0;
            while(1)
            {
                jishu++;
                double exm=ans;
                exm=sqrt(exm);
                ans=(ll)exm;
                if(ans==1)
                    break;
            }
            if(jishu<=5)
             printf("%d
",jishu);
            else
             printf("TAT
");
        }
    }
    return 0;
}