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POJ1284_Primitive Roots【欧拉函数】

2024-07-27 09:20:09   237人阅读

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;
}



POJ1284_Primitive Roots【欧拉函数】


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