hdu5752 Sqrt Bo（水）

思路：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

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
int cas=1,T;
char s[105];
int main()
{
//printf("%llu\n",1LL<<32);
//freopen("1.in","w",stdout);
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
//scanf("%d",&T);
while(scanf("%s",s)==1)
{
if(strlen(s)>15)
{
puts("TAT");
continue;
}
LL num;
sscanf(s,"%lld",&num);
if(!num || num>=(1LL<<32))
{
puts("TAT");
continue;
}
for(LL i=0,e=2;i<=5;i++,e=e*e)
{
if(num<e)
{
printf("%lld\n",i);
break;
}
//printf("%lld\n",e);
}
}
//printf("time=%.3lf\n",(double)clock()/CLOCKS_PER_SEC);
return 0;
}


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