POJ1284 Primitive Roots [欧拉函数,原根]
Primitive Roots
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 5434 | Accepted: 3072 |
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.
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
分析:
一句话题意:求原根的个数。
首先,如果知道原根的相关知识,那就可以直接上欧拉函数的板子了。关于原根的知识,请参考这里。
Code:
//It is made by HolseLee on 11th July 2018 //POJ 1284 #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #include<iostream> #include<iomanip> using namespace std; const int N=1e5+7; int n,phi[N],top,q[10007]; bool vis[N]; void ready() { phi[1]=1; for(int i=2;i<N;i++){ if(!vis[i])phi[q[++top]=i]=i-1; for(int j=1,k;j<=top&&(k=i*q[j])<N;j++){ vis[k]=true; if(i%q[j])phi[k]=phi[i]*(q[j]-1); else {phi[k]=phi[i]*q[j];break;} } } } int main() { ios::sync_with_stdio(false); ready(); while(cin>>n){ printf("%d\n",phi[n-1]);} return 0; }
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如需转载,请署名作者并附上原文链接,蒟蒻非常感激
名称:HolseLee
博客地址:www.cnblogs.com/cytus
个人邮箱:1073133650@qq.com