Poj_1284_Primitive root, poj_1284_primitive

Poj_1284_Primitive root, poj_1284_primitive
We say that integer x, 0 <x <p, is a primitive root modulo odd prime p if and only if the set {(x I 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

233179

Sample Output

10824

Method to Evaluate the original root of the modulo Prime Number:Decomposition of a prime factor, that is, a standard factorization type.

Is the original root. (For the original root of the sum, you only need to replace it)

 

Definition:Set, to make the smallest set up, called the order of the Peer Model, recorded.

Theorem:If the modulo has the original root, there is a total of original root.

Theorem:If yes, then.

Theorem:If it is a prime number, the prime number must exist in the original root, and the number of the original root of the modulo is.

 

Theorem:It is a positive integer. If the degree of the modulo is equal to, it is called an original root of the modulo.

Assume that a number is the original root for a modulo, and the result is different from each other. It can be called the original root of a modulo. In the final analysis, it is effective only when it is exponential. (Here is the prime number)

 

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>#include<cstdio>#define maxn 65550using namespace std;typedef long long ll;int phi[maxn];int a[maxn];bool vis[maxn];int val[maxn];int prime[maxn],pn=0;int main(){    int i,j;    for(i=1; i<=maxn; i++)        phi[i]=i;    for(i=2; i<=maxn; i+=2)        phi[i]/=2;    for(i=3; i<=maxn; i+=2)        if(phi[i]==i)        {            for(j=i; j<=maxn; j+=i)                phi[j]=phi[j]/i*(i-1);        }    int n;    while(~scanf("%d",&n))    {        cout<<phi[n-1]<<endl;    }}