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HDU 5901 Count primes (模板题)

题意:给求 1 - n 区间内的素数个数,n <= 1e11。

析:模板题。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e5 + 5;const int mod = 1e9 + 7;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}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;    while(scanf("%I64d", &n) == 1){        printf("%I64d\n", lehmer_pi(n));    }    return 0;}

 

HDU 5901 Count primes (模板题)