给定一个整数N ，任务是找到一个N个素数的序列，它们的总和是一个合成数。
例子：
Input: N = 5
Output: 2 3 5 7 11
2 + 3 + 5 + 7 + 11 = 28 which is composite.
Input: N = 6
Output: 3 5 7 11 13 17
方法：两个质数的总和始终是偶数，因为它们是除2以外的奇数。现在有两种情况
- 当N为偶数时，我们可以打印除2外的任何N个素数，它们的总和将始终为偶数，即，将偶数相加后的奇数将得到偶数和。
- 当N为奇数时，我们可以打印2和任何其他N – 1个质数，以确保总和为偶数。因为N – 1是偶数，所以除2以外的所有素数的和都将是偶数，然后我们将2作为第N个数字加起来，以确保总和保持偶数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAXN 100000
// To store prime numbers
vector v;
// Function to find and store
// all the primes <= n
void SieveOfEratosthenes(int n)
{
// 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 + 1];
memset(prime, true, sizeof(prime));
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 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 = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Store all the prime numbers
for (int p = 2; p <= n; p++)
if (prime[p])
v.push_back(p);
}
// Function to print the required sequence
void GetSequence(int n)
{
// If n is even then we do not include 2
// and start the sequence from the 2nd
// smallest prime else we include 2
int i = (n % 2 == 0) ? 1 : 0;
int cnt = 0;
// Print the sequence
while (cnt < n) {
cout << v[i] << " ";
i++;
cnt++;
}
}
// Driver code
int main()
{
int n = 6;
SieveOfEratosthenes(MAXN);
GetSequence(n);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
static int MAXN = 100000;
// To store prime numbers
static Vector v = new Vector();
// Function to find and store
// all the primes <= n
static void SieveOfEratosthenes(int n)
{
// 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.
boolean[] prime = new boolean[n + 1];
Arrays.fill(prime,true);
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 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 = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Store all the prime numbers
for (int p = 2; p <= n; p++)
if (prime[p])
v.add(p);
}
// Function to print the required sequence
static void GetSequence(int n)
{
// If n is even then we do not include 2
// and start the sequence from the 2nd
// smallest prime else we include 2
int i = (n % 2 == 0) ? 1 : 0;
int cnt = 0;
// Print the sequence
while (cnt < n)
{
System.out.print(v.get(i) + " ");
i++;
cnt++;
}
}
// Driver code
public static void main(String[] args)
{
int n = 6;
SieveOfEratosthenes(MAXN);
GetSequence(n);
}
}
// This code is contributed by Princi Singh python
# Python3 implementation of the approach
MAXN=100000
# To store prime numbers
v=[]
# Function to find and store
# all the primes <= n
def SieveOfEratosthenes(n):
# 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 for i in range(n + 1)]
for p in range(2,n+1):
# 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 i in range(2 * p, n + 1, p):
prime[i] = False
# Store all the prime numbers
for p in range(2, n + 1):
if (prime[p]):
v.append(p)
# Function to print the required sequence
def GetSequence(n):
# If n is even then we do not include 2
# and start the sequence from the 2nd
# smallest prime else we include 2
if n % 2 == 0:
i = 1
else:
i = 0
cnt = 0
# Print the sequence
while (cnt < n):
print(v[i],end=" ")
i += 1
cnt += 1
# Driver code
n = 6
SieveOfEratosthenes(MAXN)
GetSequence(n)
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int MAXN = 100000;
// To store prime numbers
static List v = new List();
// Function to find and store
// all the primes <= n
static void SieveOfEratosthenes(int n)
{
// 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.
Boolean[] prime = new Boolean[n + 1];
for(int i = 0; i < n + 1; i++)
prime[i] = true;
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 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 = p * p; i <= n; i += p)
prime[i] = false;
}
}
// Store all the prime numbers
for (int p = 2; p <= n; p++)
if (prime[p])
v.Add(p);
}
// Function to print the required sequence
static void GetSequence(int n)
{
// If n is even then we do not include 2
// and start the sequence from the 2nd
// smallest prime else we include 2
int i = (n % 2 == 0) ? 1 : 0;
int cnt = 0;
// Print the sequence
while (cnt < n)
{
Console.Write(v[i] + " ");
i++;
cnt++;
}
}
// Driver code
public static void Main(String[] args)
{
int n = 6;
SieveOfEratosthenes(MAXN);
GetSequence(n);
}
}
/* This code is contributed by PrinciRaj1992 */ 输出：
3 5 7 11 13 17