给定一个整数数组(小于10 ^ 6)，任务是查找在每个(k-1)个质数之后出现的所有质数的总和
即数组中的第K个素数。
例子：
Input : Array : 2, 3, 5, 7, 11 ; n=5; k=2
Output : Sum = 10
Explanation: All the elements of the array are prime. So,
the prime numbers after every K intervals are 3, 7 and their sum is 10.
Input : Array : 41, 23, 12, 17, 18, 19 ; n=6; k=2
Output : Sum = 42一个简单的方法
我们必须遍历数组，并在每个(k-1)个质数之后找到质数。这样，我们将必须检查数组的每个元素是否为质数，随着数组大小的增加，这将花费更多时间。
高效的方法
我们将创建一个筛子来存储数字是否为质数。然后，它可以用于检查O(1)时间中是否有素数。这样，我们只需要跟踪每个第K个素数并保持运行总和即可。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 1000000
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
memset(prime, true, sizeof(prime));
// 0 and 1 are not prime numbers
prime[1] = false;
prime[0] = false;
for (int p = 2; p * p <= MAX; 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 <= MAX; i += p)
prime[i] = false;
}
}
}
// compute the answer
void solve(int arr[], int n, int k)
{
// count of primes
int c = 0;
// sum of the primes
long long int sum = 0;
// traverse the array
for (int i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
cout << sum << endl;
}
// Driver code
int main()
{
// create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int arr[n] = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static final int MAX=1000000;
static boolean []prime=new boolean[MAX + 1];
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
for(int i=0;i<=MAX;i++)
prime[i]=true;
// 0 and 1 are not prime numbers
prime[1] = false;
prime[0] = false;
for (int p = 2; p * p <= MAX; 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 <= MAX; i += p)
prime[i] = false;
}
}
}
// compute the answer
static void solve(int []arr, int n, int k)
{
// count of primes
int c = 0;
// sum of the primes
long sum = 0;
// traverse the array
for (int i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
System.out.println(sum);
}
// Driver code
public static void main(String []args)
{
// create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int []arr = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
}
}Python3
# Python3 implementation of the approach
def SieveOfEratosthenes():
# 0 and 1 are not prime numbers
prime[1] = False
prime[0] = False
p = 2
while p * p <= MAX:
# If prime[p] is not changed,
# then it is a prime
if prime[p] == True:
# Update all multiples of p
for i in range(p * 2, MAX + 1, p):
prime[i] = False
p += 1
# Compute the answer
def solve(arr, n, k):
# count of primes
c = 0
# sum of the primes
Sum = 0
# Traverse the array
for i in range(0, n):
# if the number is a prime
if prime[arr[i]]:
# increase the count
c += 1
# if it is the K'th prime
if c % k == 0:
Sum += arr[i]
c = 0
print(Sum)
# Driver code
if __name__ == "__main__":
MAX = 1000000
prime = [True] * (MAX + 1)
# Create the sieve
SieveOfEratosthenes()
n, k = 5, 2
arr = [2, 3, 5, 7, 11]
solve(arr, n, k)
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
class GFG
{
static int MAX=1000000;
static bool []prime=new bool[MAX + 1];
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
for(int i=0;i<=MAX;i++)
prime[i]=true;
// 0 and 1 are not prime numbers
prime[1] = false;
prime[0] = false;
for (int p = 2; p * p <= MAX; 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 <= MAX; i += p)
prime[i] = false;
}
}
}
// compute the answer
static void solve(int []arr, int n, int k)
{
// count of primes
int c = 0;
// sum of the primes
long sum = 0;
// traverse the array
for (int i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
Console.WriteLine(sum);
}
// Driver code
public static void Main()
{
// create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int []arr = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
}
}Javascript
输出：
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