给定整数N ，任务是找到小于或等于N的半素数之和。半质数是两个质数的倍数。
例子：
Input: N = 6
Output: 10
4 and 6 are the semi primes ≤ 6
4 + 6 = 10
Input: N = 10000000
Output: 9322298311255
方法：
- 首先使用Sieve计算小于或等于N的素数，并将它们按排序顺序存储在向量中。
- 遍历素数的向量。修复其中一个素数，并开始使用此固定素数检查所有素数的乘积值。
- 由于素数是按排序顺序排列的，因此一旦我们找到积超过N的素数，那么所有剩余素数都将超过该素数。因此，在这里中断嵌套循环。
- 将所有有效对的乘积值添加到答案变量。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define ll long long int
// Vector to store the primes
vector pr;
// Create a boolean array "prime[0..n]"
bool prime[10000000 + 1];
void sieve(ll n)
{
// Initialize all prime values to be true
for (int i = 2; i <= n; i += 1) {
prime[i] = 1;
}
for (ll p = 2; (ll)p * (ll)p <= n; p++) {
// If prime[p] is not changed then it is a prime
if (prime[p] == true) {
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked
for (ll i = (ll)p * (ll)p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (ll p = 2; p <= n; p++)
if (prime[p])
pr.push_back(p);
}
// Function to return the semi-prime sum
ll SemiPrimeSum(ll N)
{
// Variable to store the sum of semi-primes
ll ans = 0;
// Iterate over the prime values
for (int i = 0; i < pr.size(); i += 1) {
for (int j = i; j < pr.size(); j += 1) {
// Break the loop once the product exceeds N
if ((ll)pr[i] * (ll)pr[j] > N)
break;
// Add valid products which are less than
// or equal to N
// each product is a semi-prime number
ans += (ll)pr[i] * (ll)pr[j];
}
}
return ans;
}
// Driver code
int main()
{
ll N = 6;
sieve(N);
cout << SemiPrimeSum(N);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Vector to store the primes
static Vector pr = new Vector<>();
// Create a boolean array "prime[0..n]"
static boolean prime[] = new boolean[10000000 + 1];
static void sieve(long n)
{
// Initialize along prime values to be true
for (int i = 2; i <= n; i += 1)
{
prime[i] = true;
}
for (int p = 2; (int)p * (int)p <= n; p++)
{
// If prime[p] is not changed then it is a prime
if (prime[p] == true)
{
// Update along multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked
for (int i = (int)p * (int)p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int p = 2; p <= n; p++)
if (prime[p])
pr.add((long)p);
}
// Function to return the semi-prime sum
static long SemiPrimeSum(long N)
{
// Variable to store the sum of semi-primes
long ans = 0;
// Iterate over the prime values
for (int i = 0; i < pr.size(); i += 1)
{
for (int j = i; j < pr.size(); j += 1)
{
// Break the loop once the product exceeds N
if ((long)pr.get(i) * (long)pr.get(j) > N)
break;
// Add valid products which are less than
// or equal to N
// each product is a semi-prime number
ans += (long)pr.get(i) * (long)pr.get(j);
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
long N = 6;
sieve(N);
System.out.println(SemiPrimeSum(N));
}
}
// This code is contributed by Princi Singh Python3
# Python3 implementation of the approach
# Vector to store the primes
pr=[]
# Create a boolean array "prime[0..n]"
prime = [1 for i in range(10000000 + 1)]
def sieve(n):
for p in range(2, n):
if p * p > n:
break
# If prime[p] is not changed then it is a prime
if (prime[p] == True):
# Update amultiples of p greater than or
# equal to the square of it
# numbers which are multiple of p and are
# less than p^2 are already been marked
for i in range(2 * p, n + 1, p):
prime[i] = False
# Praprime numbers
for p in range(2, n + 1):
if (prime[p]):
pr.append(p)
# Function to return the semi-prime sum
def SemiPrimeSum(N):
# Variable to store the sum of semi-primes
ans = 0
# Iterate over the prime values
for i in range(len(pr)):
for j in range(i,len(pr)):
# Break the loop once the product exceeds N
if (pr[i] * pr[j] > N):
break
# Add valid products which are less than
# or equal to N
# each product is a semi-prime number
ans += pr[i] * pr[j]
return ans
# Driver code
N = 6
sieve(N)
print(SemiPrimeSum(N))
# This code is contributed by mohit kumar 29C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Vector to store the primes
static List pr = new List();
// Create a boolean array "prime[0..n]"
static bool []prime = new bool[10000000 + 1];
static void sieve(long n)
{
// Initialize along prime values to be true
for (int i = 2; i <= n; i += 1)
{
prime[i] = true;
}
for (int p = 2; (int)p * (int)p <= n; p++)
{
// If prime[p] is not changed then it is a prime
if (prime[p] == true)
{
// Update along multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked
for (int i = (int)p * (int)p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for (int p = 2; p <= n; p++)
if (prime[p])
pr.Add((long)p);
}
// Function to return the semi-prime sum
static long SemiPrimeSum(long N)
{
// Variable to store the sum of semi-primes
long ans = 0;
// Iterate over the prime values
for (int i = 0; i < pr.Count; i += 1)
{
for (int j = i; j < pr.Count; j += 1)
{
// Break the loop once the product exceeds N
if ((long)pr[i] * (long)pr[j] > N)
break;
// Add valid products which are less than
// or equal to N
// each product is a semi-prime number
ans += (long)pr[i] * (long)pr[j];
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
long N = 6;
sieve(N);
Console.WriteLine(SemiPrimeSum(N));
}
}
// This code is contributed by Rajput-Ji 输出：
10