POJ1284_Primitive Roots【欧拉函数】

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Primitive Roots

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 3032Accepted: 1734

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

题目大意：p是奇素数,如果{x^i % p | 1 <= i <= p - 1} = {1,2,...,p-1}，则称x是p的原根。
给出一个p，问它的原根有多少个。

思路：

{x^i% p | 1 <= i <= p - 1} = {1,2,...,p-1} 等价于
{x^i%(p-1) | 1 <= i <= p - 1} = {0,1,2,...,p-2}，

即{x^1，x^2，x^3，…，x^(p-1)}为p的完全剩余系等价于

若x与p-1互质（gcd(x, p-1) = 1），则{x^0，x^1，x^2，…，x^(p-2)}为(p-1)的完全剩余系

下边证明：

如果x^i != x^j (mod p-1)，那么 x*x^i != x*x^j (mod p-1)，则gcd(x, p-1) != 0，与上边相悖。

则x^i = x^j(mod p-1)。

根据费马定理和欧拉定理知：i = φ(p-1)。

关于欧拉函数、费马定理和欧拉定理参考另一篇欧拉函数的题解：

http://blog.csdn.net/lianai911/article/details/40114675

参考博文：http://www.cnblogs.com/lnever/p/3969117.html

顺便膜拜推理出来的大神

#include <stdio.h>#include <math.h>int Euler(int n){    int i, ret = n;    for(i = 2; i <= sqrt(1.0*n); i++)    {        if(n % i == 0)        {            n /= i;            ret = ret - ret/i;        }        while(n % i == 0)            n /= i;    }    if(n > 1)    {        ret = ret - ret/n;    }    return ret;}int main(){    int p;    while(~scanf("%d",&p))    {        printf("%d\n",Euler(p-1));        //如果p为素数，Euler(Euler(p)) == Euler(p-1)    }    return 0;}

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