hdu5752 Sqrt Bo（水）

最新推荐文章于 2019-01-19 19:07:07 发布

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本文介绍了一个通过迭代求解特定数学函数的方法，该函数定义为 f(n)=⌊n−−√⌋，并给出了一个C++实现示例。博文详细探讨了如何在限定时间内找到使该函数迭代结果为1的最小输入值。

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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