POJ 1284 Primitive Roots

Primitive Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5481		Accepted: 3101

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

我觉得丧心病狂的是读了很久题才看懂意思（英语太渣了。。）再一看题目，原根。。好吧，就当进一步理解原根了吧。
原根定义：设m是正整数，a是整数，若a模m的阶等于φ(m)，则称a为模m的一个原根。（其中φ(m)表示m的欧拉函数） 

假设一个数g是P的原根，那么g^i mod P的结果两两不同,且有 1< g< P, 0< i < P,归根到底就是g^(P-1) = 1 (mod P)当且仅当指数为P-1的时候成立.(这里P是素数). 

简单来说，g^i mod p ≠ g^j mod p （p为素数） 

其中i≠j且i, j介于1至(p-1)之间 

则g为p的原根。
用两次欧拉函数就ok了。

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include<iostream>
#include <cmath>
#include<string>
#define ll long long 
#define dscan(a) scanf("%d",&a)
#define mem(a,b) memset(a,b,sizeof a)
using namespace std;
#define MAXL 1105
#define Endl endl
#define maxn 1000005
ll x,y;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
int euler(int n)
{
     int i;
     int res = n,a = n;
     for(i = 2;i*i <= a; ++i)
     {
         if(a%i == 0)
         {
             res -= res/i; 
             while(a%i == 0) a/=i;
         }
     }
     if(a > 1) res -= res/a;
     return res;
}
int main()
{
    int n;
    while(cin>>n)
    {
       cout<<euler(euler(n))<<endl;
    }
    return 0;
}