1 条题解
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0
欧拉筛法
#include <bits/stdc++.h> using namespace std; vector<int> Prime(int n){ // 求解n以内(含n)的素数 bool flag[n + 1]; // 标记数组,flag[i]==0表示i为素数,flag[i]==1表示i为合数 memset(flag, 0, sizeof(flag)); vector<int> prime; int cnt = 0; // 素数个数 for (int i = 2; i <= n; ++i) { if (!flag[i]) { prime.push_back(i); // 将i加入素数表 cnt++; } for (int j = 0; j < cnt; ++j) { // 保证每个合数只会被它的最小质因数筛去 if (i * prime[j] > n) break; flag[i * prime[j]] = 1; if (i % prime[j] == 0) break; } } return prime; } int main(int argc, char const *argv[]) { int n; while(1) { printf("请输入n,将输出n以内(含n)的素数:"); scanf("%d", &n); if(n < 0) break; vector<int> prime = Prime(n); int cnt = prime.size(); printf("一共有%d个素数:\n", cnt); for(int i = 0; i < cnt; i++) { printf("%3d ", prime[i]); if(i % 10 == 9) puts(""); } puts("\n"); } return 0; }
Meisell-Lehmer 模板 这个是我偶然看到的网上的,自己也没仔细看
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int N = 5e6 + 2; bool np[N]; int prime[N], pi[N]; int getprime() { int cnt = 0; np[0] = np[1] = true; pi[0] = pi[1] = 0; for(int i = 2; i < N; ++i) { if(!np[i]) prime[++cnt] = i; pi[i] = cnt; for(int j = 1; j <= cnt && i * prime[j] < N; ++j) { np[i * prime[j]] = true; if(i % prime[j] == 0) break; } } return cnt; } const int M = 7; const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17; int phi[PM + 1][M + 1], sz[M + 1]; void init() { getprime(); sz[0] = 1; for(int i = 0; i <= PM; ++i) phi[i][0] = i; for(int i = 1; i <= M; ++i) { sz[i] = prime[i] * sz[i - 1]; for(int j = 1; j <= PM; ++j) { phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1]; } } } int sqrt2(LL x) { LL r = (LL)sqrt(x - 0.1); while(r * r <= x) ++r; return int(r - 1); } int sqrt3(LL x) { LL r = (LL)cbrt(x - 0.1); while(r * r * r <= x) ++r; return int(r - 1); } LL getphi(LL x, int s) { if(s == 0) return x; if(s <= M) return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s]; if(x <= prime[s]*prime[s]) return pi[x] - s + 1; if(x <= prime[s]*prime[s]*prime[s] && x < N) { int s2x = pi[sqrt2(x)]; LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2; for(int i = s + 1; i <= s2x; ++i) { ans += pi[x / prime[i]]; } return ans; } return getphi(x, s - 1) - getphi(x / prime[s], s - 1); } LL getpi(LL x) { if(x < N) return pi[x]; LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1; for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) { ans -= getpi(x / prime[i]) - i + 1; } return ans; } LL lehmer_pi(LL x) { if(x < N) return pi[x]; int a = (int)lehmer_pi(sqrt2(sqrt2(x))); int b = (int)lehmer_pi(sqrt2(x)); int c = (int)lehmer_pi(sqrt3(x)); LL sum = getphi(x, a) + LL(b + a - 2) * (b - a + 1) / 2; for (int i = a + 1; i <= b; i++) { LL w = x / prime[i]; sum -= lehmer_pi(w); if (i > c) continue; LL lim = lehmer_pi(sqrt2(w)); for (int j = i; j <= lim; j++) { sum -= lehmer_pi(w / prime[j]) - (j - 1); } } return sum; } int main() { init(); LL n; cin >> n ; cout << lehmer_pi(n) << endl; return 0; }
- 1
信息
- ID
- 451
- 时间
- 2000ms
- 内存
- 125MiB
- 难度
- 9
- 标签
- 递交数
- 11
- 已通过
- 5
- 上传者