HDU 5901-Count primes（1e11以内素数的个数）

电玩女神 2022-09-26 04:59 184阅读 0赞

# Count primes #

**Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)  
Total Submission(s): 971 Accepted Submission(s): 520**

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][2016 ACM_ICPC Asia Regional Shenyang Online]

Recommend

wange2014 | We have carefully selected several similar problems for you: [5900][] [5899][] [5898][] [5897][] [5896][]

## 题目意思： ##

输入N，求N（N最大取到 1e11）以内素数的个数。

## 解题思路： ##

智商和刷题量的完全碾压…以下代码全是我从网上搜罗的，膜一下大神们…Orz

AC代码①

![Center][]

#include <bits/stdc++.h>
    #define ll long long
    using namespace std;
    ll f[340000],g[340000],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)!=EOF){
            init();
            cout<<f[1]<<endl;
        }
        return 0;
    }

AC代码②

![Center 1][]

//Meisell-Lehmer  
    #include<cstdio>  
    #include<cmath>  
    using namespace std;  
    #define LL long long  
    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;  
        while(~scanf("%lld",&n))  
        {  
            printf("%lld\n",lehmer_pi(n));  
        }  
        return 0;  
    }

超内存代码③（超了一万）：

#include <stdio.h>
    #include <stdlib.h>
    #include <stack>
    #include <math.h>
    #include<bits/stdc++.h>
    #include<malloc.h>
    
    #define XPREVN		(22*1000*1000)				/**/
    #define XSQRTPREVN	(5*1000)
    #define XPREVPI		(2*1000*1000)				/* PI(XPREVN) : 3957809 */
    
    int		*X_small_prime,  X_pi_small_prime;
    int		*X_pi_prev_N, *X_prev_V;
    
    #define XPREVDM		(7)
    #define XPREVQ		(2*3*5*7*11*13*17)
    #define XPREVPHQ	(1*2*4*6*10*12*16)
    
    typedef long long	LLong;
    
    stack st;
    LLong phiX_p( LLong x, LLong a )
    {
        if( a == XPREVDM )
            return x / XPREVQ * XPREVPHQ + X_prev_V[ x % XPREVQ ];
        if( x < X_small_prime[a-1] )
            return 1;
        return phiX_p( x, a - 1 ) - phiX_p( x / X_small_prime[a-1], a - 1 );
    }
    LLong phiX_p( LLong x, LLong a )
    {
        long long ans;
        if( a == XPREVDM ) ans=x / XPREVQ * XPREVPHQ + X_prev_V[ x % XPREVQ ];
        if( x < X_small_prime[a-1] ) ans=1;
        return ans;
    }
    
    
    LLong phiX( LLong X )
    {
        LLong cubeRootX, max, len3, len2, s, i, k;
        LLong sum, p;
        if( X <= XPREVN )
            return X_pi_prev_N[X];
    
        max   = (LLong)(pow( (double)X, 2./3 ) + 2);
        cubeRootX = (LLong)(pow( (double)X, 1./3 ) + .5);
        if( cubeRootX*cubeRootX*cubeRootX > X )
            --cubeRootX;
        len3 = X_pi_prev_N[cubeRootX];
    
        sum = 0;
        s = 0;
        k = max -1 ;
        for( i = len3; ; ++i )
        {
            p = X_small_prime[i];
            if( p * p > X ) break;
    
            s += X_pi_prev_N[k] - X_pi_prev_N[(int)((double)X/p)];
            k = (int)((double)X/p);
    
            sum += s;
        }
        len2 = X_pi_prev_N[p-1];
    
        sum = (len2-len3)*X_pi_prev_N[max-1] - sum;
        sum = len3 * (len3-1)/2 - len2 * (len2-1) / 2 + sum;
    
        return phiX_p( X, len3 ) - sum + len3 - 1;
    }
    
    void initSmallPrime()
    {
        char *mark = (char*)calloc(1,XPREVN/2+1);
        int i, j;
    
        X_small_prime = (int*)calloc(sizeof(int),XPREVPI );
        X_pi_prev_N   = (int*)calloc(sizeof(int),XPREVN+1);
        X_prev_V	  = (int*)calloc(sizeof(int),XPREVQ+1);
    
        for( i = 3; i <= XSQRTPREVN; i += 2 )
        {
            if( mark[i/2] ) continue;
            for( j = i*i/2; j <= XPREVN/2; j+=i )
                mark[j] = 1;
        }
        X_small_prime[0] = 2;
        X_pi_small_prime = 1;
        X_pi_prev_N[0] = X_pi_prev_N[1] = 0;
        X_pi_prev_N[2] = 1;
        for( i = 3; i <= XPREVN; i += 2 )
        {
            if( mark[i/2] )
            {
                X_pi_prev_N[i+1] = X_pi_prev_N[i] = X_pi_prev_N[i-1];
            }
            else
            {
                X_pi_prev_N[i+1] = X_pi_prev_N[i] = 1 + X_pi_prev_N[i-1];
                X_small_prime[X_pi_small_prime++] = i;
            }
        }
        free(mark);
    
        for( i = 0; i < XPREVQ; ++i )
            X_prev_V[i] = i;
        for( i = 1; i <= XPREVDM; ++i )
            for( j = XPREVQ-1; j >= 0; --j )
                X_prev_V[j] -= X_prev_V[ j/X_small_prime[i-1] ];
    }
    
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/x/read.txt","r",stdin);
        freopen("G:/x/out.txt","w",stdout);
    #endif
        LLong	x, r;
        initSmallPrime();
    
        while(~scanf( "%I64d", &x ))
        {
            r = phiX( x );
            printf("%I64d\n",r);
        }
        free(X_small_prime);
        free(X_pi_prev_N);
        free(X_prev_V);
        return 0;
    }

[2016 ACM_ICPC Asia Regional Shenyang Online]: http://acm.hdu.edu.cn/search.php?field=problem&key=2016+ACM%2FICPC+Asia+Regional+Shenyang+Online&source=1&searchmode=source
[5900]: http://acm.hdu.edu.cn/showproblem.php?pid=5900
[5899]: http://acm.hdu.edu.cn/showproblem.php?pid=5899
[5898]: http://acm.hdu.edu.cn/showproblem.php?pid=5898
[5897]: http://acm.hdu.edu.cn/showproblem.php?pid=5897
[5896]: http://acm.hdu.edu.cn/showproblem.php?pid=5896
[Center]: /images/20220717/584be8854b6a4c4cbaa22431712ef16d.png
[Center 1]: /images/20220717/9ef3cbc0d7fc4b7eb83ad76d5ffdbffe.png