HDU5752 Sqrt Bo（2016多校训练）

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1173    Accepted Submission(s): 528

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

 

题意：给出一个字符串，若开五次根号能开到1，则输出开方次数，否则输出“TAT”。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int main()
{
    char str[1000];
    ll i,len;
    ll n = 0,ans = 0;
    while(~scanf("%s",str))
    {
        len = strlen(str);
        if(len>10)
        {
            i = 0;
        }
        else
        {
            i = 1;
            for(int j = 0; str[j] && j < 11; j++)
                n = n*10 + str[j]-'0';
            //cout << n << endl;
            while(n != 1)
            {
                n = (long long)sqrt((double)n);
                ans++;
                if(ans > 5)
                    break;
            }
        }
        n = 0;
        //cout << i << ' ' << ans << endl;
        if(i && ans <= 5)
            cout << ans << endl;
        else
            cout << "TAT\n";
        ans = 0;
        i = 0;
    }
    return 0;
}