给定一个整数k和一个整数arr (小于10 ^ 6)的数组，任务是找到数组中第k个素数的和。
例子：
Input: arr[] = {2, 3, 5, 7, 11}, k = 2
Output: 10
All the elements of the array are prime. So, the prime numbers after every K (i.e. 2) interval are 3, 7 and their sum is 10.
Input: arr[] = {11, 13, 15, 17, 19}, k = 2
Output: 32
简单方法：遍历数组，找到数组中的每个第K个质数，然后计算运行总和。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。
高效方法：创建一个筛子，该筛子将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 SumOfKthPrimes(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 arr[] = { 2, 3, 5, 7, 11 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
SumOfKthPrimes(arr, n, k);
return 0;
} Java
// Java implementation of the approach
public class GFG {
static 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 < prime.length; 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 SumOfKthPrimes(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 arr[] = {2, 3, 5, 7, 11};
int n = arr.length;
int k = 2;
SumOfKthPrimes(arr, n, k);
}
}Python3
# Python3 implementation of the approach
MAX = 100000;
prime = [True] * (MAX + 1);
def 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.
# 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]):
# Update all multiples of p
i = p * 2;
while(i <= MAX):
prime[i] = False;
i += p;
p += 1;
# compute the answer
def SumOfKthPrimes(arr, n, k):
# count of primes
c = 0;
# sum of the primes
sum = 0;
# traverse the array
for i in range(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
# create the sieve
SieveOfEratosthenes();
arr = [ 2, 3, 5, 7, 11 ];
n = len(arr);
k = 2;
SumOfKthPrimes(arr, n, k);
# This code is contributed by mitsC#
// C# implementation of the approach
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.
// 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] == false)
{
// Update all multiples of p
for (int i = p * 2;
i <= MAX; i += p)
prime[i] = true;
}
}
}
// compute the answer
static void SumOfKthPrimes(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.Console.WriteLine(sum);
}
// Driver code
static void Main()
{
// create the sieve
SieveOfEratosthenes();
int[] arr = new int[] { 2, 3, 5, 7, 11 };
int n = arr.Length;
int k = 2;
SumOfKthPrimes(arr, n, k);
}
}
// This code is contributed by mitsPHP
输出：
10