Count primes

题目意思就是求1到n之间的素数个数 n<=1e11。

思路就是: 不会啊, 先存两份模板.

顺便存存高手代码。暂时没看懂，不过好短，感觉很厉害。

#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
",pcf::Lehmer(n));
        //printf("%lld
", res);
    }
    return 0;
}

模板二代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<vector>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<time.h>

using namespace std;
typedef long long LL;
const int INF=2e9+1e8;
const int MOD=1e9+7;
const int MAXSIZE=1e6+5;
const double eps=0.0000000001;
void fre()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
}
#define memst(a,b) memset(a,b,sizeof(a))
#define fr(i,a,n) for(int i=a;i<n;i++)


LL f[320000],g[320000],n;

void init()
{
    LL i,j,m;
    for(m=1; m*m<=n; ++m) f[m]=n/m-1;
    for(i=1; i<=m; ++i)   g[i]=i-1;
    for(i=2; i<=m; ++i)
    {
        if(g[i]==g[i-1])continue;
        for(j=1; j<=min(m-1,n/i/i); ++j)
        {
            if(i*j<m)f[j]-=f[i*j]-g[i-1];
            else f[j]-=g[n/i/j]-g[i-1];
        }
        for(j=m; j>=i*i; --j)g[j]-=g[j/i]-g[i-1];
    }
}
int main()
{
    while(scanf("%I64d",&n)&&n)
    {
        init();
        cout<<f[1]<<endl;
    }
    return 0;

}

--> 点击打开链接题解