POJ1284---Primitive Roots(求原根个数, 欧拉函数)
2015-06-06 08:59:28         来源：Alex

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

23
31
79

Sample Output

10
8
24

Source

``````/*************************************************************************
> File Name: POJ1284.cpp
> Author: ALex
> Mail: www.2cto.com
> Created Time: 2015年06月04日 星期四 16时04分32秒
************************************************************************/

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair  PLL;

int phi[100000];
int minDiv[100000];

void geteuler() {
for (int i = 1; i <= 70000; ++i) {
minDiv[i] = i;
}
for (int i = 2; i <= 70000; ++i) {
if (minDiv[i] == i) {
if (70000 / i < i) {
break;
}
for (int j = i * i; j <= 70000; j += i) {
minDiv[j] = i;
}
}
}
phi[1] = 1;
for (int i = 2; i <= 70000; ++i) {
phi[i] = phi[i / minDiv[i]];
if ((i / minDiv[i]) % minDiv[i]) {
phi[i] *= (minDiv[i] - 1);
}
else {
phi[i] *= minDiv[i];
}
}
}

int main() {
geteuler();
int n;
while (cin >> n) {
cout << phi[n - 1] << endl;
}
return 0;
}``````