poj1284 Primitive Roots

最新推荐文章于 2020-07-03 11:17:06 发布

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最新推荐文章于 2020-07-03 11:17:06 发布

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原文链接：http://www.cnblogs.com/zbtrs/p/8082404.html

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本文介绍了一种求解特定范围内奇质数p的原根个数的方法，并给出了一段C++实现代码，该算法利用欧拉函数φ来确定原根的数量。

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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 ⁱ 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[70000],prime[70000],tot,vis[70000];

void init()
{
    phi[1] = 1;
    for (int i = 2; i <= 70000; i++)
    {
        if (!vis[i])
        {
            prime[++tot] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= tot; j++)
        {
            int t = prime[j] * i;
            if (t > 70000)
                break;
            vis[t] = 1;
            if (i % prime[j] == 0)
            {
                phi[t] = phi[i] * prime[j];
                break;
            }
            phi[t] = phi[i] * (prime[j] - 1);
        }
    }
}

int main()
{
    init();
    while (scanf("%d",&n) != EOF)
        printf("%d\n",phi[n - 1]);

}