joj 2330: Math

ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE1s8192K39554Standard

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

621250180245361

Sample Output

2367820/*

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. }