joj 2330: Math

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	8192K	395	54	Standard

Given a nonnegative integer a, and a positive integer N, we define:
f(a, 1) = a
f(a, k) = f(a, k – 1) * f(a, k – 1) % N, k > 1
There may or may not exist some positive integer k satisfying f(a, k) = 0.
Your task is, given a positive integer N, to determine how many a (0 <= a <= N) there are, such that for some positive integer k, f(a, k) = 0.

Input

The input contains an integer T on the first line, indicating the number of test cases. Each test case contains only one positive integer N (1 <= N <= 1000000000) on a line.

Output

For each test case, output the answer on a single line.

Sample Input

6
2
12
50
180
245
361

Sample Output

2
3
6
7
8
20
/*

最悲剧的一道题，一开用cin,cout来输入输出数据，一直TLE，各种剪枝，各种预处理，都TLE，后来改正scanf,printf坑爹的居然过了。。。

于是果断以后再也不用cin，cout了。

  我的代码不算快，有师兄0.05s过的,我的0.09s,有机会膜拜下。

*/

1. #include <cstdio>
2. #include <iostream>
3. #include <memory>
4. #include <cmath>
5. using namespace std;
6. int prime [ 10000 ];
7. const int maxn= 40000;
8. int search ( int n, int f )
9. {
10.     int m= sqrt (n +0.5 );
11.    for ( int i=f ; prime [i ]<=m &&i< 8000; i++ )
12.     if (n%prime [i ]== 0 ) return prime [i ];
13.    return n;
14. }
15. int main ( )
16. {
17. int c = 0;
18. bool vis [maxn ] ;
19.     memset (vis , 1 , sizeof (vis ) );
20.     for ( int i= 2; i< 220 ; i++ )
21.      if (vis [i ] )
22.       {
23.         for ( int j=i*i; j<maxn ; j+=i )
24.         vis [j ]= 0;
25.       }//筛素数
26.     for ( int i= 2 ; i<maxn ;i++ )
27.      if (vis [i ] )
28.      prime [c++ ]=i;//将筛出的素数存入数组中
29. /*   for(int i=0 ; i<50 ; i++)
30.     cout<<prime[i]<<" ";*/
31. int t;
32. cin>>t;
33. while (t-- )
34. {
35.     int  n,i= 1;
36.     scanf ( "%d",&n );
37.    int   nn=n;
38.     int  ans = 1;
39.     int j= 0;
40.     while (n!= 1 )
41.     {
42.      int d=search (n,j++ );
43.      ans*=d;
44.      while (n%d== 0 )n=n/d;
45.     }//找出n的所有素因子（不重复）并保存其积
46.     ans=nn/ans + 1;
47.     printf ( "%d/n",ans );
48. }
49. return 0;
50. }