POJ-1284 Primitive Roots(欧拉函数)

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)

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include "algorithm"
using namespace std;
typedef long long LL;
const int MAX=65540;
int n;
int phi[MAX];
void euler(){
    int i,j;
    for (i=1;i<MAX;i++) phi[i]=i;
    for (i=2;i<MAX;i+=2) phi[i]/=2;
    for (i=3;i<MAX;i+=2)
     if (phi[i]==i)
      for (j=i;j<MAX;j+=i)
       phi[j]=phi[j]/i*(i-1);
}
int main(){
    freopen ("roots.in","r",stdin);
    freopen ("roots.out","w",stdout);
    euler();int i,j;
    while (~scanf("%d",&n)){
        printf("%d
",phi[n-1]);
    }
    return 0;
}