Sqrt Bo
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 423 Accepted Submission(s): 188
Problem Description
Let's define the function $f(n)=\lfloor \sqrt{n}\rfloor$.
Bo wanted to know the minimum number $y$ which satisfies $f^y(n)=1$.
note:$f^1(n)=f(n),f^y(n)=f(f^{y-1}(n))$
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 $n(n<10^{100})$.
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
Source
Recommend
wange2014
题意：给你一个数 问你需要不断开根多少次结果等于1 超过5次输出 TAT
题解：由于有5次的这个限制，所以尝试寻找分界点。
很容易发现是2^{32}，所以我们先比较输入的数字是否比这个大，然后再暴力开根
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
//#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
#define mod 1000000007
#define PI acos(-1.0)
using namespace std;
char a[];
ll ans;
int main()
{
while(scanf("%s",a)!=EOF){
ans=;
int len=strlen(a);
if(len>||(len==&&a[]==''))
printf("TAT\n");
else
{
for(int i=;i<len;i++)
ans=ans*+a[i]-'';
int jishu=;
while()
{
jishu++;
double exm=ans;
exm=sqrt(exm);
ans=(ll)exm;
if(ans==)
break;
}
if(jishu<=)
printf("%d\n",jishu);
else
printf("TAT\n");
}
}
return ;
}