HDU 5901-Count primes（1e11以内素数的个数）

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

 

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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

#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

//Meisell-Lehmer
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
const int N = 5e6 + 2;
bool np
;
int prime
, pi
;
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;
}