给定数字N。找到从1到的几乎素数的数量
。如果一个数字恰好具有两个截然不同的素数，则称其为“几乎”。
注意：数字可以具有任意数量的非质数，但应恰好具有两个质数。
例子：
Input : N = 10
Output : 2
Explanation : 6, 10 are such numbers.
Input : N = 21
Output : 8
一种有效的解决方案是使用Eratosthenes筛子查找素数。并找到不同的素数来计算小于N的数。
请参考：几乎素数
下面是上述方法的实现：
C++
// CPP program to count almost prime numbers
// from 1 to n
#include
using namespace std;
#define N 100005
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[N];
void SieveOfEratosthenes()
{
memset(prime, true, sizeof(prime));
prime[1] = false;
for (int p = 2; p * p < N; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i < N; i += p)
prime[i] = false;
}
}
}
// Function to count almost prime numbers
// from 1 to n
int almostPrimes(int n)
{
// to store required answer
int ans = 0;
// 6 is first almost prime number
for (int i = 6; i <= n; i++) {
// to count prime factors
int c = 0;
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
// if it is perfect square
if (j * j == i) {
if (prime[j])
c++;
}
else {
if (prime[j])
c++;
if (prime[i / j])
c++;
}
}
}
// if I is almost prime number
if (c == 2)
ans++;
}
return ans;
}
// Driver code
int main()
{
SieveOfEratosthenes();
int n = 21;
cout << almostPrimes(n);
return 0;
} Java
// Java program to count almost prime numbers
// from 1 to n
import java.io.*;
class GFG {
static int N = 100005;
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
static boolean prime[] = new boolean[N];
static void SieveOfEratosthenes()
{
for(int i=0;iPython 3
# Python 3 program to count almost
# prime numbers
# from 1 to n
# from math import everything
from math import *
N = 100005
# Create a boolean array "prime[0..n]"
# and initialize all entries it as true.
# A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True] * N
def SieveOfEratosthenes() :
prime[1] = False
for p in range(2, int(sqrt(N))) :
# If prime[p] is not changed, then
# it is a prime
if prime[p] == True :
# Update all multiples of p
for i in range(2*p, N, p) :
prime[i] = False
# Function to count almost prime numbers
# from 1 to n
def almostPrimes(n) :
# to store required answer
ans = 0
# 6 is first almost prime number
for i in range(6, n + 1) :
# to count prime factors
c = 0
for j in range(2, int(sqrt(i)) + 1) :
# if it is perfect square
if i % j == 0 :
if j * j == i :
if prime[j] :
c += 1
else :
if prime[j] :
c += 1
if prime[i // j] :
c += 1
# if I is almost prime number
if c == 2 :
ans += 1
return ans
# Driver Code
if __name__ == "__main__" :
SieveOfEratosthenes()
n = 21
print(almostPrimes(n))
# This code is contributed by ANKITRAI1C#
// C# program to count almost
// prime numbers from 1 to n
using System;
class GFG
{
static int N = 100005;
// Create a boolean array "prime[0..n]"
// and initialize all entries it as
// true. A value in prime[i] will finally
// be false if i is Not a prime, else true.
static bool []prime = new bool[N];
static void SieveOfEratosthenes()
{
for(int i = 0; i < N; i++)
prime[i] = true;
prime[1] = false;
for (int p = 2; p * p < N; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2; i < N; i += p)
prime[i] = false;
}
}
}
// Function to count almost
// prime numbers from 1 to n
static int almostPrimes(int n)
{
// to store required answer
int ans = 0;
// 6 is first almost prime number
for (int i = 6; i <= n; i++)
{
// to count prime factors
int c = 0;
for (int j = 2; j * j <= i; j++)
{
if (i % j == 0)
{
// if it is perfect square
if (j * j == i)
{
if (prime[j])
c++;
}
else
{
if (prime[j])
c++;
if (prime[i / j])
c++;
}
}
}
// if I is almost prime number
if (c == 2)
ans++;
}
return ans;
}
// Driver code
public static void Main ()
{
SieveOfEratosthenes();
int n = 21;
Console.WriteLine( almostPrimes(n));
}
}
// This code is contributed
// by inder_vermaPHP
= $j * $j; $j++)
{
// if it is perfect square
if ($i % $j == 0)
{
if ($j * $j == $i)
{
if ($prime[$j])
$c += 1;
}
else
{
if ($prime[$j])
$c += 1;
if ($prime[($i / $j)])
$c += 1;
}
}
}
// if I is almost prime number
if ($c == 2)
$ans += 1;
}
return $ans;
}
// Driver Code
SieveOfEratosthenes();
$n = 21;
print(almostPrimes($n));
// This code is contributed by mits
?>输出：
8