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
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 }
时间: 2024-10-14 21:30:58