hdoj 5752

Let's define the function f(n)=⌊n−−√⌋f(n)=⌊n⌋. 

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

note:f1(n)=f(n),fy(n)=f(fy−1(n))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.

InputThis problem has multi test cases(no more than 120120). 

Each test case contains a non-negative integer n(n<10100)n(n<10100).OutputFor each test case print a integer - the answer yyor a string "TAT" - Bo can't solve this problem.Sample Input

233
233333333333333333333333333333333333333333333333333333333

Sample Output

3
TAT

/*
要在五次内开跟达到1，第一次要在4以内，第二次在16以内，
    第三次256，第四次65536，第五次4294967296，所以超过10位的都是TAT*/

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;

char ch[110];

long long i;
const long long num=4294967296-1;

int main()
{
    while(~scanf("%s",ch))
    {
        int r=strlen(ch);
        int l=0;
        while(ch[l]=='0') l++;
        if(r-l>10)
        {
            printf("TAT
");
            continue ;
        }
        i=0;
        while(l<r)
        {
            i=i*10+(ch[l]-'0');
            l++;
        }
        //cout<<i<<endl;
        if(i>num||i==0)
        {
            printf("TAT
");
        }
        else
        {
            int ii=0;
            while(i!=1)
            {
                i=(long long )sqrt(i);
                //cout<<i<<endl;
                ii++;
            }
            printf("%d
",ii);
        }
    }
}

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
#define MAXN 10000000
char str [MAXN];
int main()
{
    while(~scanf("%s",str)){
        int len=strlen(str);
        if(len>10){printf("TAT");continue;}
        else{
            long long  n=0;bool flag=0;
            for(int i=0;i<len;i++) n=n*10+str[i]-'0';
            for(int i=1;i<=5;i++){
                n=sqrt(n);
                if(n==1){flag=1;printf("%d
",i);break;}
            }
            if(flag==0) printf("TAT");
        }
    }
    return 0;
}