Primitive Roots

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 3984

 

Accepted: 2373

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7. 
Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p. 

Input

Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.

Output

For each p, print a single number that gives the number of primitive roots in a single line.

Sample Input

23
31
79

Sample Output

10
8
24

Source

​​贾怡@pku​​

直接看了题解，求φ(n-1)

1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=65540;
13 int n;
14 int phi[MAX];
15 void euler(){
16     int i,j;
17     for (i=1;i<MAX;i++) phi[i]=i;
18     for (i=2;i<MAX;i+=2) phi[i]/=2;
19     for (i=3;i<MAX;i+=2)
20      if (phi[i]==i)
21       for (j=i;j<MAX;j+=i)
22        phi[j]=phi[j]/i*(i-1);
23 }
24 int main(){
25     freopen ("roots.in","r",stdin);
26     freopen ("roots.out","w",stdout);
27     euler();int i,j;
28     while (~scanf("%d",&n)){
29         printf("%d\n",phi[n-1]);
30     }
31     return 0;
32

请看上一题题解