Meissel Lehmer Algorithm 求前n个数中素数个数 【模板】
Count primes
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2719 Accepted Submission(s): 1386
Problem Description
Easy question! Calculate how many primes between [1...n]!
Input
Each line contain one integer n(1 <= n <= 1e11).Process to end of file.
Output
For each case, output the number of primes in interval [1...n]
Sample Input
2 3 10
Sample Output
1 2 4
Source
2016 ACM/ICPC Asia Regional Shenyang Online
#include <bits/stdtr1c++.h>
#define MAXN 100
#define MAXM 10001
#define MAXP 40000
#define MAX 400000
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))
using namespace std;
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;
}
}
long long solve(long long n){
int i, j, k, l;
long long x, y, res = 0;
for (i = 0; i < pcf::len; i++){
x = pcf::primes[i], y = n / x;
if ((x * x) > n) break;
res += (pcf::Lehmer(y) - pcf::Lehmer(x));
}
for (i = 0; i < pcf::len; i++){
x = pcf::primes[i];
if ((x * x * x) > n) break;
res++;
}
return res;
}
int main(){
pcf::init();
long long n, res;
while (scanf("%lld", &n) != EOF){
//res = solve(n);
printf("%lld\n",pcf::Lehmer(n));
//printf("%lld\n", res);
}
return 0;
}