HDU 5752 Sqrt Bo(多校3--1001)

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 493    Accepted Submission(s): 227

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

233233333333333333333333333333333333333333333333333333333333

 

Sample Output

3TAT

题解：由于有5的的这个限制，所以尝试寻找分界点，很容易发现分界点是2^32,所以我们判断这个数字是否比2^32要大，如果大的话直接输出‘TAT‘，否则暴力开根即可

#include <algorithm>#include <iostream>#include <numeric>#include <cstring>#include <iomanip>#include <string>#include <vector>#include <cstdio>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const LL Max = (1LL<<32) - 1;const double esp = 1e-6;const double PI = 3.1415926535898;const int INF = 0x3f3f3f3f;using namespace std;char arr[120];int main(){    while(~scanf("%s",arr)){        LL num = 0;        bool f = false;        for(int i=0;arr[i]!='\0';i++){            num = num * 10 + (arr[i] - '0');            if(num > Max){                f = true;                break;            }        }        if(f || num == 0)            printf("TAT\n");        else{            int ans = 0;            while(num > 1){                ans += 1;                num = floor(sqrt(num));            }            printf("%d\n",ans);        }    }    return 0;}