hdu 5752 Sqrt Bo 水题

Sqrt Bo

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5752

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

Hint

题意

给你一个数，问你最少开多少次，可以变成1。

如果超过五次还没有变成1，就输出TAT

题解：

象征性猜一猜，五次的根号不会很大。

1->3->15->255->65535->4294967295

那就最大就是4294967295了，超过就直接输出TAT好了

代码

#include<bits/stdc++.h>
using namespace std;
string s;
void solve(){
    if(s.size()>18){
        printf("TAT\n");
        return;
    }
    long long n = 0;
    for(int i=0;i<s.size();i++){
        n = n*10+s[i]-'0';
    }
    if(n==1){
        cout<<"0"<<endl;
        return;
    }
    for(int i=1;i<=5;i++){
        n = sqrt(n);
        if(n==1){
            cout<<i<<endl;
            return;
        }
    }
    cout<<"TAT"<<endl;
}
int main(){
    while(cin>>s)solve();
    return 0;
}