poj1284 Primitive Roots
2024-08-25 16:43:32
Primitive Roots
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 4775 | Accepted: 2827 |
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
Source
题目大意:求n的原根个数.
分析:结论题,n的原根个数=φ(φ(n)).
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; int n,phi[],prime[],tot,vis[]; void init()
{
phi[] = ;
for (int i = ; i <= ; i++)
{
if (!vis[i])
{
prime[++tot] = i;
phi[i] = i - ;
}
for (int j = ; j <= tot; j++)
{
int t = prime[j] * i;
if (t > )
break;
vis[t] = ;
if (i % prime[j] == )
{
phi[t] = phi[i] * prime[j];
break;
}
phi[t] = phi[i] * (prime[j] - );
}
}
} int main()
{
init();
while (scanf("%d",&n) != EOF)
printf("%d\n",phi[n - ]); }
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