poj1284 Primitive Roots

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4775		Accepted: 2827

Description

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (x 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

题目大意：求n的原根个数.

分析：结论题，n的原根个数=φ(φ(n)).

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

int n,phi[],prime[],tot,vis[];

void init()

{

    phi[] = ;

    for (int i = ; i <= ; i++)

    {

        if (!vis[i])

        {

            prime[++tot] = i;

            phi[i] = i - ;

        }

        for (int j = ; j <= tot; j++)

        {

            int t = prime[j] * i;

            if (t > )

                break;

            vis[t] = ;

            if (i % prime[j] == )

            {

                phi[t] = phi[i] * prime[j];

                break;

            }

            phi[t] = phi[i] * (prime[j] - );

        }

    }

}

int main()

{

    init();

    while (scanf("%d",&n) != EOF)

        printf("%d\n",phi[n - ]);

}