POJ1284:Primitive Roots(欧拉函数的应用,奇素数的原根)

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2159		Accepted: 1194

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

MYCode:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define MAX 65550
int prime[MAX];
bool flag[MAX];
int ct;
void pre()
{
    memset(flag,0,sizeof(flag));
    int i,j;
    flag[1]=1;
    ct=0;
    for(i=2;i*i<=MAX;i++)
    {
        if(!flag[i])
        {
            prime[ct++]=i;
            for(j=2*i;j<=MAX;j+=i)
            {
                flag[j]=1;
            }
        }
    }
}
int euler(int x)
{
    int res=x;
    int i;
    for(i=0;prime[i]*prime[i]<=x && i<ct;i++)
    {
        if(x%prime[i]==0)
        {
            res=res*1.0/prime[i]*(prime[i]-1);
            while(x%prime[i]==0)x/=prime[i];
        }
    }
    if(x>1)res=res/x*(x-1);
    return res;
}
int main()
{
    int p;
    pre();
    while(scanf("%d",&p)!=EOF)
    {
        int ans=euler(p-1);
        printf("%d\n",ans);
    }
}
//

求奇素数的原根:

结论:root=euler(euler(p))=euler(p-1)

euler函数的应用.

如何证明呢?

待定.