【SGU】SGU每日练1·Coprimes【数论】欧拉函数

先介绍欧拉函数及其相关定理吧：

1.欧拉函数：对于任意一个数X，将其分解质因数为X=(P1^b1)*(P2^b2)*(P3^b3)......

则小于X的与X互质的数的个数N为N = X * (1 - 1 / p1) * (1 - 1 / p2).......

显然对于质数p,Euler(p) = p - 1；

2.一个数的所有质因子之和是euler(n)*n/2；

3.a^Euler(n) % n = 1，这里可以用模运算来证明；

具体实现：

1.先筛素数

2.从2开始遍历素数，能整除则乘，然后将之除尽

附上代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#pragma warning(disable:4996)

#define Zero(a) memset(a, 0, sizeof(a))
#define Neg(a)  memset(a, -1, sizeof(a))
#define All(a) a.begin(), a.end()
#define PB push_back
#define repf(i,a,b) for(i = a;i <= b; i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,n,1
#define ld rt << 1
#define rd rt << 1 | 1
#define ll long long
#define MAXN 100005
#define mod 10007
using namespace std;

int prime[MAXN];
int isNotPrime[MAXN];
void get_prime(){
    int num_prime = 0;
    isNotPrime[0] = isNotPrime[1] = 1;
    for (int i = 2; i < MAXN; i++){
        if (!isNotPrime[i])
            prime[num_prime++] = i;
        for (int j = 0; j < num_prime && i * prime[j] < MAXN; j++)
        {
            isNotPrime[i * prime[j]] = 1;
            if (!(i % prime[j]))
                break;
        }
    }
}
vector <int >ok;
int main(){
    get_prime();
    //cout << prime[0] << endl;
    int n;
    while (~scanf("%d", &n)){
        ok.clear();
        int ans = n;
        for (int i = 0; prime[i] <= n; ++i){
            if (n % prime[i] == 0){
                ok.push_back(prime[i]);
                while (n % prime[i] == 0) n /= prime[i];
            }
        }
        //cout << ok.size() << endl;
        for (int i = 0; i < ok.size(); ++i){
            //cout << ok[i] << endl;
            ans *= (ok[i] - 1);
            ans /= ok[i];
        }
        printf("%d\n", ans);
    }
    return 0;
}