POJ1284 Primitive Roots【原根】

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4361		Accepted: 2568

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

问题链接：POJ1284 Primitive Roots

问题简述：参见上文。

问题分析：这个问题的题目是原根，是求奇素数的原根的个数，筛选打表实现。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* POJ1284 Primitive Roots */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 65536;

int euler[N+1];

void phi()
{
    for(int i=1; i<=N; i++)
        euler[i] = i;
    for(int i=2; i<=N; i+=2)
        euler[i] /= 2;
    for(int i=3; i<=N; i++) {
        if(euler[i] == i) { // 素数，用该素数筛
            for(int j=i; j<=N; j+=i)
                euler[j] = euler[j] / i * (i - 1);
        }
    }
}

int main()
{
    // 欧拉函数打表
    phi();

    int p;
    while(scanf("%d", &p) != EOF)
        printf("%d
", euler[p - 1]);

    return 0;
}