hdoj 5752

Let‘s define the function f(n)=?n??√?f(n)=?n?.

Bo wanted to know the minimum number yy which satisfies fy(n)=1fy(n)=1.

note:f1(n)=f(n),fy(n)=f(fy?1(n))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.

InputThis problem has multi test cases(no more than 120120).

Each test case contains a non-negative integer n(n<10100)n(n<10100).OutputFor each test case print a integer - the answer yyor a string "TAT" - Bo can‘t solve this problem.Sample Input

233
233333333333333333333333333333333333333333333333333333333

Sample Output

3
TAT

 1 /*
 2 要在五次内开跟达到1，第一次要在4以内，第二次在16以内，
 3     第三次256，第四次65536，第五次4294967296，所以超过10位的都是TAT*/
 4
 5 #include<cstdio>
 6 #include<iostream>
 7 #include<cstring>
 8 #include<queue>
 9 #include<cmath>
10 using namespace std;
11
12 char ch[110];
13
14 long long i;
15 const long long num=4294967296-1;
16
17 int main()
18 {
19     while(~scanf("%s",ch))
20     {
21         int r=strlen(ch);
22         int l=0;
23         while(ch[l]==‘0‘) l++;
24         if(r-l>10)
25         {
26             printf("TAT\n");
27             continue ;
28         }
29         i=0;
30         while(l<r)
31         {
32             i=i*10+(ch[l]-‘0‘);
33             l++;
34         }
35         //cout<<i<<endl;
36         if(i>num||i==0)
37         {
38             printf("TAT\n");
39         }
40         else
41         {
42             int ii=0;
43             while(i!=1)
44             {
45                 i=(long long )sqrt(i);
46                 //cout<<i<<endl;
47                 ii++;
48             }
49             printf("%d\n",ii);
50         }
51     }
52 }

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <string.h>
 5 using namespace std;
 6 #define MAXN 10000000
 7 char str [MAXN];
 8 int main()
 9 {
10     while(~scanf("%s",str)){
11         int len=strlen(str);
12         if(len>10){printf("TAT");continue;}
13         else{
14             long long  n=0;bool flag=0;
15             for(int i=0;i<len;i++) n=n*10+str[i]-‘0‘;
16             for(int i=1;i<=5;i++){
17                 n=sqrt(n);
18                 if(n==1){flag=1;printf("%d\n",i);break;}
19             }
20             if(flag==0) printf("TAT");
21         }
22     }
23     return 0;
24 }