POJ 1284 Primitive Roots

素数原根计数

最新推荐文章于 2024-02-08 22:43:13 发布

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最新推荐文章于 2024-02-08 22:43:13 发布

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Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4062		Accepted: 2420

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

#include<iostream>
#include<cstring>
using namespace std;
int euler(int n)
{
    int i,ans=n;
    for(i=2; i*i<=n; i++)
        if(n%i==0)
        {
            ans=ans-ans/i;
            while(n%i==0)
                n/=i;//把该素因子全部约掉
            //while(n%i==0);
        }
    if(n>1)
        ans=ans-ans/n;
    return ans;
}
int main()
{
    int n;
    while(cin>>n)
    {
        cout<<euler(n-1)<<endl;
    }
    return 0;
}