Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4505		Accepted: 2652

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

 1 //2017-08-04
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 int phi(int n){
10     int ans = n;
11     for(int i = 2; i*i <= n; i++){
12         if(n%i==0){
13             ans -= ans/i;
14             while(n%i==0)
15                 n /= i;
16         }
17     }
18     if(n > 1)ans = ans - ans/n;
19     return ans;
20 }
21 
22 int main()
23 {
24     int num;
25     while(scanf("%d", &num)!=EOF){
26         printf("%d\n", phi(num-1));
27     }
28 
29     return 0;
30 }

POJ1284(SummerTrainingDay04-K 原根)

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