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HDU5752 Sqrt Bo(2016多校训练)

时间:2016-08-02 01:06:08      阅读:236      评论:0      收藏:0      [点我收藏+]

Sqrt Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1173    Accepted Submission(s): 528


Problem Description
 
Let‘s define the function f(n)=n技术分享技术分享技术分享技术分享技术分享技术分享 .

Bo wanted to know the minimum number y技术分享 which satisfies f技术分享y技术分享(n)=1技术分享 .

note:f技术分享1技术分享(n)=f(n),f技术分享y技术分享(n)=f(f技术分享y1技术分享(n))技术分享

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技术分享 ).

Each test case contains a non-negative integer n(n<10技术分享100技术分享)技术分享 .
 

 

Output
 
For each test case print a integer - the answer y技术分享 or a string "TAT" - Bo can‘t solve this problem.
 

 

Sample Input
 
233 233333333333333333333333333333333333333333333333333333333
 

 

Sample Output
 
3 TAT
 
题意:给出一个字符串,若开五次根号能开到1,则输出开方次数,否则输出“TAT”。
 
技术分享
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string>
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 typedef long double ld;
10 
11 int main()
12 {
13     char str[1000];
14     ll i,len;
15     ll n = 0,ans = 0;
16     while(~scanf("%s",str))
17     {
18         len = strlen(str);
19         if(len>10)
20         {
21             i = 0;
22         }
23         else
24         {
25             i = 1;
26             for(int j = 0; str[j] && j < 11; j++)
27                 n = n*10 + str[j]-0;
28             //cout << n << endl;
29             while(n != 1)
30             {
31                 n = (long long)sqrt((double)n);
32                 ans++;
33                 if(ans > 5)
34                     break;
35             }
36         }
37         n = 0;
38         //cout << i << ‘ ‘ << ans << endl;
39         if(i && ans <= 5)
40             cout << ans << endl;
41         else
42             cout << "TAT\n";
43         ans = 0;
44         i = 0;
45     }
46     return 0;
47 }
View Code

 

HDU5752 Sqrt Bo(2016多校训练)

原文:http://www.cnblogs.com/Mino521/p/5727728.html

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