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 $f(n)=\lfloor \sqrt{n}\rfloor$.

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[];
 ll ans;
 int main()
 {
       while(scanf("%s",a)!=EOF){
         ans=;
         int len=strlen(a);
         if(len>||(len==&&a[]==''))
          printf("TAT\n");
         else
         {
             for(int i=;i<len;i++)
                 ans=ans*+a[i]-'';
             int jishu=;
             while()
             {
                 jishu++;
                 double exm=ans;
                 exm=sqrt(exm);
                 ans=(ll)exm;
                 if(ans==)
                     break;
             }
             if(jishu<=)
              printf("%d\n",jishu);
             else
              printf("TAT\n");
         }
     }
     return ;
 }