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 ;

}

巴特西

hdu 5752 Sqrt Bo

Sqrt Bo

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