hdu-5752 Sqrt Bo(水题)

题目链接：

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 131072/131072 K (Java/Others)

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:

 

问这个数开几次平方得到1;如果超过五的话就TAT;

 

思路：

 

开五次方还没变成1的数大于十位了吧;

然后把这个数转成long long 看开几次平方变成1;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  unsigned long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e7+10;
const int maxn=1000+10;
const double eps=1e-8;


char s[110];

int main()
{
        while(scanf("%s",s)!=EOF)
        {
            int len=strlen(s);
            if(len>10)cout<<"TAT"<<"
";
            else
            {
                LL x=0;
                for(int i=0;i<len;i++)
                {
                    x=x*10;
                    x=x+s[i]-'0';
                }
                if(x==0){cout<<"TAT
";continue;}
                int ans=0;
                while(ans<=10)
                {
                    x=floor(sqrt(x));
                    ans++;
                    if(x==1)break;
                }
                if(ans>5)cout<<"TAT"<<"
";
                else cout<<ans<<"
";
            }
        }
        return 0;
}