给定K个数组，其中第一个数组包含第一个质数，第二个数组包含接下来的2个质数，第三个数组包含接下来的3个质数，依此类推。任务是在第K个数组中找到素数之和。
例子：
Input: K = 3
Output: 31
arr1[] = {2}
arr[] = {3, 5}
arr[] = {7, 11, 13}
7 + 11 + 13 = 31
Input: K = 2
Output: 8
方法：可以使用Eratosthenes筛子来查找直至所需元素的所有素数。从1到K – 1的数组中素数的计数将为cnt = 1 + 2 + 3 +…+(K – 1)=(K *(K – 1))/ 2 。现在，从筛子阵列的第(cnt + 1)个素数开始，开始加所有素数，直到精确地添加了K个素数，然后打印总和。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 1000000
// To store whether a number is prime or not
bool prime[MAX];
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes()
{
// 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.
for (int i = 0; i < MAX; i++)
prime[i] = true;
for (int p = 2; p * p < MAX; p++) {
// If prime[p] is not changed then it is a prime
if (prime[p]) {
// 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 < MAX; i += p)
prime[i] = false;
}
}
}
// Function to return the sum of
// primes in the Kth array
int sumPrime(int k)
{
// Update vector v to store all the
// prime numbers upto MAX
SieveOfEratosthenes();
vector v;
for (int i = 2; i < MAX; i++) {
if (prime[i])
v.push_back(i);
}
// To store the sum of primes
// in the kth array
int sum = 0;
// Count of primes which are in
// the arrays from 1 to k - 1
int skip = (k * (k - 1)) / 2;
// k is the number of primes
// in the kth array
while (k > 0) {
sum += v[skip];
skip++;
// A prime has been
// added to the sum
k--;
}
return sum;
}
// Driver code
int main()
{
int k = 3;
cout << sumPrime(k);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static int MAX = 1000000;
// To store whether a number is prime or not
static boolean []prime = new boolean[MAX];
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
// 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.
for (int i = 0; i < MAX; i++)
prime[i] = true;
for (int p = 2; p * p < MAX; p++)
{
// If prime[p] is not changed
// then it is a prime
if (prime[p])
{
// 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 < MAX; i += p)
prime[i] = false;
}
}
}
// Function to return the sum of
// primes in the Kth array
static int sumPrime(int k)
{
// Update vector v to store all the
// prime numbers upto MAX
SieveOfEratosthenes();
Vector v = new Vector<>();
for (int i = 2; i < MAX; i++)
{
if (prime[i])
v.add(i);
}
// To store the sum of primes
// in the kth array
int sum = 0;
// Count of primes which are in
// the arrays from 1 to k - 1
int skip = (k * (k - 1)) / 2;
// k is the number of primes
// in the kth array
while (k > 0)
{
sum += v.get(skip);
skip++;
// A prime has been
// added to the sum
k--;
}
return sum;
}
// Driver code
public static void main(String[] args)
{
int k = 3;
System.out.println(sumPrime(k));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation of the approach
from math import sqrt
MAX = 1000000
# 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] * MAX
# Function for Sieve of Eratosthenes
def SieveOfEratosthenes() :
for p in range(2, int(sqrt(MAX)) + 1) :
# If prime[p] is not changed
# then it is a prime
if (prime[p]) :
# 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(p * p, MAX, p) :
prime[i] = False;
# Function to return the sum of
# primes in the Kth array
def sumPrime(k) :
# Update vector v to store all the
# prime numbers upto MAX
SieveOfEratosthenes();
v = [];
for i in range(2, MAX) :
if (prime[i]) :
v.append(i);
# To store the sum of primes
# in the kth array
sum = 0;
# Count of primes which are in
# the arrays from 1 to k - 1
skip = (k * (k - 1)) // 2;
# k is the number of primes
# in the kth array
while (k > 0) :
sum += v[skip];
skip += 1;
# A prime has been
# added to the sum
k -= 1;
return sum;
# Driver code
if __name__ == "__main__" :
k = 3;
print(sumPrime(k));
# This code is contributed by AnkitRai01C#
// C# mplementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int MAX = 1000000;
// To store whether a number is prime or not
static bool []prime = new bool[MAX];
// Function for Sieve of Eratosthenes
static void SieveOfEratosthenes()
{
// 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.
for (int i = 0; i < MAX; i++)
prime[i] = true;
for (int p = 2; p * p < MAX; p++)
{
// If prime[p] is not changed
// then it is a prime
if (prime[p])
{
// 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 < MAX; i += p)
prime[i] = false;
}
}
}
// Function to return the sum of
// primes in the Kth array
static int sumPrime(int k)
{
// Update vector v to store all the
// prime numbers upto MAX
SieveOfEratosthenes();
List v = new List();
for (int i = 2; i < MAX; i++)
{
if (prime[i])
v.Add(i);
}
// To store the sum of primes
// in the kth array
int sum = 0;
// Count of primes which are in
// the arrays from 1 to k - 1
int skip = (k * (k - 1)) / 2;
// k is the number of primes
// in the kth array
while (k > 0)
{
sum += v[skip];
skip++;
// A prime has been
// added to the sum
k--;
}
return sum;
}
// Driver code
public static void Main(String[] args)
{
int k = 3;
Console.WriteLine(sumPrime(k));
}
}
// This code is contributed by PrinciRaj1992 输出：
31