HDU 5752 Sqrt Bo【枚举,大水题】

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HDU 5752 Sqrt Bo【枚举,大水题】

angel_kitty 2017-07-11 09:00:00 浏览402
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Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2221    Accepted Submission(s): 882


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(fy1(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
 

 

Author
绍兴一中
 

 

Source
题意:给你一个数,对这个数进行连续开方,这个数能在5次以内开方(每次开方都是向下取整),使得最后结果等于1,输出开方次数,否则输出TAT
分析:看起来这道题怪吓人的,10^100次,lfh跟我说这道题很难,要用字符串啥玩意来着,然后算什么2^32的数还是啥的,弱弱菜鸡就随手扔了一个代码
枚举一波连续开5次根的数,然后。。。。竟然AC了,旁边的lfh懵逼了:这也行?(弱弱的说他好像Wa了好几发的样子),此题数据是10^100,建议开long double型
其次是。。。。。我特别嫌弃这种要连续输入的题,每次害得我找错误找半天QAQ!
下面给出AC代码:
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long double ll;
 4 inline ll gcd(ll x)
 5 {
 6     int k,t;
 7     int flag=0;
 8     for(k=1;k<=5;k++)
 9     {
10         x=(long long)sqrt(x);
11         if(x==1)
12         {
13             flag=1;
14             t=k;
15             break;
16         }
17     }
18     if(flag)
19         return t;
20     return 0;
21 }
22 int main()
23 {
24     ll n;
25     while(cin>>n)
26     {
27     if(!gcd(n))
28         cout<<"TAT"<<endl;
29     else cout<<gcd(n)<<endl;
30     }
31     return 0;
32 }

 

 

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