标签:limit set scan using for res mod RoCE nta
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2719 Accepted Submission(s): 1386
#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; }
Meissel Lehmer Algorithm 求前n个数中素数个数 【模板】
标签:limit set scan using for res mod RoCE nta
原文地址:https://www.cnblogs.com/zxyqzy/p/10223769.html