[AcWing 874] 筛法求欧拉函数
作者:互联网
复杂度 $ O(n) $
总体复杂度 $ 10^{6} $
点击查看代码
#include<iostream>
using namespace std;
const int N = 1e6 + 10;
typedef long long LL;
int primes[N], cnt;
int eulers[N];
bool st[N];
void get_eulers(int n)
{
eulers[1] = 1;
for (int i = 2; i <= n; i ++) {
if (!st[i]) {
primes[cnt ++] = i;
eulers[i] = i - 1;
}
for (int j = 0; primes[j] <= n / i; j ++) {
st[primes[j] * i] = true;
if (i % primes[j] == 0) {
eulers[primes[j] * i] = eulers[i] * primes[j];
break;
}
eulers[primes[j] * i] = eulers[i] * (primes[j] - 1);
}
}
}
int main()
{
int n;
cin >> n;
get_eulers(n);
LL res = 0;
for (int i = 1; i <= n; i ++) res += eulers[i];
cout << res << endl;
return 0;
}
- 记 $ primes[j] $ 为 $ P_j $,分以下三种情况求欧拉函数:
① $ i $ 是质数,$ 1 $ ~ $ i - 1 $ 每个数都与 $ i $ 互质,$ \phi(i) = i - 1 $;
② $ P_j $ 是 $ i $ 的质因子,由于 $ i = P_1^{\alpha_1} \times P_2^{\alpha_2} \times \cdots \times P_k^{\alpha_k} $ ,$ P_j $ 必是 $ P_1 $ ~ $ P_k $ 其中的一项,那么 $ P_j \times i = P_1^{\alpha_1} \times P_2^{\alpha_2} \times \cdots \times P_k^{\alpha_k} $ ,由欧拉函数公式,$ \phi(P_j \times i) = P_j \times i \times (1 - \frac{1}{P_1}) \times (1 - \frac{1}{P_2}) \times \cdots \times (1 - \frac{k}{P_k}) = P_j \times \phi(i) $
③ $ P_j $ 不是 $ i $ 的质因子,由于 $ i = P_1^{\alpha_1} \times P_2^{\alpha_2} \times \cdots \times P_k^{\alpha_k} $ ,$ P_j $ 必不在 $ P_1 $ ~ $ P_k $ 中,那么 $ P_j \times i = P_j \times P_1^{\alpha_1} \times P_2^{\alpha_2} \times \cdots \times P_k^{\alpha_k} $ ,由欧拉函数公式,$ \phi(P_j \times i) = P_j \times (1 - \frac{1}{P_j}) \times i \times (1 - \frac{1}{P_1}) \times (1 - \frac{1}{P_2}) \times \cdots \times (1 - \frac{k}{P_k}) = P_j \times (1 - \frac{1}{P_j}) \times \phi(i) $
标签:phi,frac,筛法,int,874,times,cdots,alpha,AcWing 来源: https://www.cnblogs.com/wKingYu/p/16248654.html