A

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A - Primitive Roots

POJ - 1284

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

Sample Output

题目链接：https://cn.vjudge.net/contest/157463#problem/A

原根问题，这一类题有求最小原根，原根的个数，具体怎么推得我忘记了，好像有什么欧拉函数？积性函数？GCD？我只记住结论了。

代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int dp[100000];int prime[100000];bool vis[100000];void Init(){    dp[1]=1;    int top=0;    for(int i=2;i<100000;i++){        if(!vis[i]){            prime[++top]=i;            dp[i]=i-1;        }        for(int j=1;j<=top;j++){            if(i*prime[j]>=100000)                break;            vis[i*prime[j]]=1;            if(i%prime[j]==0){                dp[i*prime[j]]=dp[i]*prime[j];//只记住了这个结论                break;            }            else{                dp[i*prime[j]]=dp[i]*(prime[j]-1);                //cout<<i*prime[j]<<dp[i*prime[j]]<<endl;            }        }    }}int main(){    Init();    int n;    while(~scanf("%d",&n)){        printf("%d\n",dp[n-1]);    }    return 0;}

刚才上网上收了几个大牛的博客，51nod上有个题，1135，是求最小原根的

http://blog.csdn.net/huayunhualuo/article/details/51095717

http://blog.csdn.net/corncsd/article/details/33414871?locationNum=2&fps=1

http://blog.csdn.net/wahaha1_/article/details/8078753

http://blog.csdn.net/tsaid/article/details/7296307

这个是51nod的最小原根

http://blog.csdn.net/u013486414/article/details/47781857

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