2017sdut省赛选拔（1）--poj1284(原根问题+欧拉回路)

http://poj.org/problem?id=1284

Primitive Roots

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4271 Accepted: 2503

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

233179

Sample Output

10824

题意比较简单。问题是通过这个题意转到原根问题上来。如果了解过原根的一些定义定理的话也不难联想。关于一些原根的问题建议去一些百科或博客看一下。当读一遍后套上定理这个题就成了。我是用的欧拉函数来求解的，根据定理预处理一下，然后直接套就行了。关于欧拉函数是什么区百度。

///求原根的个数#include <iostream>#include <cstdio>#include<cstring>#include<cmath>using namespace std;int pri[71000];int a[71000];int vis[71000];void phi(){    memset(vis,0,sizeof(vis));    for(int i=2; i<=sqrt(70000+0.5); i++)    {        if(!vis[i])        {            for(int j=i*i; j<=70000; j+=i)vis[j]=1;        }    }    int top=0;    for(int i=2; i<=70000; i++)    {        if(!vis[i]){pri[top++]=i;}    }    for(int j=2; j<=65536; j++)    {        long long num=1;        int x=j;        int sum=x;        for(int i=0; i<top&&pri[i]<=x; i++)        {            if(x%pri[i]==0)            {                sum=sum*(pri[i]-1);                num=num*pri[i];                while(x%pri[i]==0)                {                    x=x/pri[i];                }            }        }        a[j]=sum/num;        //cout<<a[j]<<endl;    }}int main(){    phi();///求欧拉函数    int p;    while(~scanf("%d",&p))    {        printf("%d\n",a[a[p]]);///因为是奇素数，所以输出a[p-1]也行    }    return 0;}