hdu 5752 Sqrt Bo

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 980    Accepted Submission(s): 452

Problem Description

Let's define the function f(n)=⌊n√⌋.

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

note:f1(n)=f(n),fy(n)=f(fy−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<10100).

 

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

 

Author

绍兴一中

 

Source

2016 Multi-University Training Contest 3

 

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string>
#include<string.h>
using namespace std;
int main()
{

    string s6="";
    string s5="";
    string s4="";
    string s3="";
    string s2="";
    string s1="";
    string a;
    while(cin>>a)
    {
        if((a.length()==s6.length()&&a>=s6)||(a.length()>s6.length()))
        {
            printf("TAT\n");
        }
        else if((a.length()==s5.length()&&a>=s5)||(a.length()>s5.length())){printf("5\n");}
        else if((a.length()==s4.length()&&a>=s4)||(a.length()>s4.length())){printf("4\n");}
        else if((a.length()==s3.length()&&a>=s3)||(a.length()>s3.length())){printf("3\n");}
        else if((a.length()==s2.length()&&a>=s2)||(a.length()>s2.length())){printf("2\n");}
        else if((a.length()==s1.length()&&a>=s1)||(a.length()>s1.length())){printf("1\n");}
        else if(a=="") printf("0\n");
        else if(a=="") printf("TAT\n");
    }
    return ;

}