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:
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Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 #include<iostream> 7 #include<iomanip> 8 using namespace std; 9 const int N=1e5+7; 10 int n,phi[N],top,q[10007]; 11 bool vis[N]; 12 void ready() 13 { 14 phi[1]=1; 15 for(int i=2;i<N;i++){ 16 if(!vis[i])phi[q[++top]=i]=i-1; 17 for(int j=1,k;j<=top&&(k=i*q[j])<N;j++){ 18 vis[k]=true; 19 if(i%q[j])phi[k]=phi[i]*(q[j]-1); 20 else {phi[k]=phi[i]*q[j];break;} 21 } 22 } 23 } 24 int main() 25 { 26 ios::sync_with_stdio(false); 27 ready(); 28 while(cin>>n){ 29 printf("%d\n",phi[n-1]);} 30 return 0; 31 }