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Meissel-Lehmer算法(求N以内素数个数)

程序员文章站 2024-03-14 21:08:38
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比欧拉筛快。

#define MAXN 100    // pre-calc max n for phi(m, n)
#define MAXM 10010 // pre-calc max m for phi(m, n)
#define MAXP 40000 // max primes counter
#define MAX 400010    // max prime
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31)))) 
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))
 
namespace pcf {
	long long dp[MAXN][MAXM];
	unsigned int ar[(MAX >> 6) + 5] = { 0 };
	int len = 0, primes[MAXP], counter[MAX];
 
	void Sieve() {
		setbit(ar, 0), setbit(ar, 1);
		for (int i = 3; (i * i) < MAX; i++, i++) {
			if (!chkbit(ar, i)) {
				int k = i << 1;
				for (int j = (i * i); j < MAX; j += k) setbit(ar, j);
			}
		}
 
		for (int i = 1; i < MAX; i++) {
			counter[i] = counter[i - 1];
			if (isprime(i)) primes[len++] = i, counter[i]++;
		}
	}
 
	void init() {
		Sieve();
		for (int n = 0; n < MAXN; n++) {
			for (int m = 0; m < MAXM; m++) {
				if (!n) dp[n][m] = m;
				else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
			}
		}
	}
 
	long long phi(long long m, int n) {
		if (n == 0) return m;
		if (primes[n - 1] >= m) return 1;
		if (m < MAXM && n < MAXN) return dp[n][m];
		return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
	}
 
	long long Lehmer(long long m) {
		if (m < MAX) return counter[m];
 
		long long w, res = 0;
		int i, a, s, c, x, y;
		s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
		a = counter[y], res = phi(m, a) + a - 1;
		for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
		return res;
	}
}
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