给定一个由n个整数和数字K组成的数组arr [] 。任务是计算具有精确的K个质数的子数组的数量。
例子：
Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 4
Explanation:
Since total number of prime number in the array are 2. So the 4 subarray with 2 prime number are:
1. {2, 3}
2. {1, 2, 3}
3. {2, 3, 4}
4. {1, 2, 3, 4}
Input: arr[] = {2, 4, 5}, K = 3
Output: 0
Explanation:
Since total number of prime number in the array are 2 which is less than K(K = 3).
So there is no such subarray with K primes.
方法：
- 遍历给定的数组arr []并检查元素是否为质数。
- 如果当前元素为质数，则将该索引的数组的值更改为1，否则将该索引的值更改为0。
- 现在，给定的数组将转换为Binary Array。
- 使用本文讨论的方法，在上述二进制数组中找到总和等于K的子数组的计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// A utility function to check if
// the number n is prime or not
bool isPrime(int n)
{
int i;
// Base Cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// Check to skip middle five
// numbers in below loop
if (n % 2 == 0 || n % 3 == 0) {
return false;
}
for (i = 5; i * i <= n; i += 6) {
// If n is divisible by i & i+2
// then it is not prime
if (n % i == 0
|| n % (i + 2) == 0) {
return false;
}
}
return true;
}
// Function to find number of subarrays
// with sum exactly equal to k
int findSubarraySum(int arr[], int n, int K)
{
// STL map to store number of subarrays
// starting from index zero having
// particular value of sum.
unordered_map prevSum;
int res = 0;
// To store the sum of element traverse
// so far
int currsum = 0;
for (int i = 0; i < n; i++) {
// Add current element to currsum
currsum += arr[i];
// If currsum = K, then a new
// subarray is found
if (currsum == K) {
res++;
}
// If currsum > K then find the
// no. of subarrays with sum
// currsum - K and exclude those
// subarrays
if (prevSum.find(currsum - K)
!= prevSum.end())
res += (prevSum[currsum - K]);
// Add currsum to count of
// different values of sum
prevSum[currsum]++;
}
// Return the final result
return res;
}
// Function to count the subarray with K primes
void countSubarray(int arr[], int n, int K)
{
// Update the array element
for (int i = 0; i < n; i++) {
// If current element is prime
// then update the arr[i] to 1
if (isPrime(arr[i])) {
arr[i] = 1;
}
// Else change arr[i] to 0
else {
arr[i] = 0;
}
}
// Function Call
cout << findSubarraySum(arr, n, K);
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
countSubarray(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG {
// A utility function to check if
// the number n is prime or not
static boolean isPrime(int n) {
int i;
// Base Cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// Check to skip middle five
// numbers in below loop
if (n % 2 == 0 || n % 3 == 0) {
return false;
}
for (i = 5; i * i <= n; i += 6) {
// If n is divisible by i & i+2
// then it is not prime
if (n % i == 0 || n % (i + 2) == 0) {
return false;
}
}
return true;
}
// Function to find number of subarrays
// with sum exactly equal to k
static int findSubarraySum(int arr[], int n, int K)
{
// STL map to store number of subarrays
// starting from index zero having
// particular value of sum.
HashMap prevSum =
new HashMap();
int res = 0;
// To store the sum of element traverse
// so far
int currsum = 0;
for (int i = 0; i < n; i++) {
// Add current element to currsum
currsum += arr[i];
// If currsum = K, then a new
// subarray is found
if (currsum == K) {
res++;
}
// If currsum > K then find the
// no. of subarrays with sum
// currsum - K and exclude those
// subarrays
if (prevSum.containsKey(currsum - K)) {
res += (prevSum.get(currsum - K));
}
// Add currsum to count of
// different values of sum
if (prevSum.containsKey(currsum))
prevSum.put(currsum, prevSum.get(currsum) + 1);
else
prevSum.put(currsum, 1);
}
// Return the final result
return res;
}
// Function to count the subarray with K primes
static void countSubarray(int arr[], int n, int K) {
// Update the array element
for (int i = 0; i < n; i++) {
// If current element is prime
// then update the arr[i] to 1
if (isPrime(arr[i])) {
arr[i] = 1;
}
// Else change arr[i] to 0
else {
arr[i] = 0;
}
}
// Function Call
System.out.print(findSubarraySum(arr, n, K));
}
// Driver Code
public static void main(String[] args) {
int arr[] = { 1, 2, 3, 4 };
int K = 2;
int N = arr.length;
// Function Call
countSubarray(arr, N, K);
}
}
// This code contributed by Rajput-Ji Python3
# Python3 program for the above approach
from math import sqrt
# A utility function to check if
# the number n is prime or not
def isPrime(n):
# Base Cases
if (n <= 1):
return False
if (n <= 3):
return True
# Check to skip middle five
# numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return False
for i in range(5,int(sqrt(n))+1,6):
# If n is divisible by i & i+2
# then it is not prime
if (n % i == 0 or n % (i + 2) == 0):
return False
return True
# Function to find number of subarrays
# with sum exactly equal to k
def findSubarraySum(arr,n,K):
# STL map to store number of subarrays
# starting from index zero having
# particular value of sum.
prevSum = {i:0 for i in range(100)}
res = 0
# To store the sum of element traverse
# so far
currsum = 0
for i in range(n):
# Add current element to currsum
currsum += arr[i]
# If currsum = K, then a new
# subarray is found
if (currsum == K):
res += 1
# If currsum > K then find the
# no. of subarrays with sum
# currsum - K and exclude those
# subarrays
if (currsum - K) in prevSum:
res += (prevSum[currsum - K])
# Add currsum to count of
# different values of sum
prevSum[currsum] += 1
# Return the final result
return res
# Function to count the subarray with K primes
def countSubarray(arr,n,K):
# Update the array element
for i in range(n):
# If current element is prime
# then update the arr[i] to 1
if (isPrime(arr[i])):
arr[i] = 1
# Else change arr[i] to 0
else:
arr[i] = 0
# Function Call
print(findSubarraySum(arr, n, K))
# Driver Code
if __name__ == '__main__':
arr = [1, 2, 3, 4]
K = 2
N = len(arr)
# Function Call
countSubarray(arr, N, K)
# This code is contributed by Surendra_GangwarC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// A utility function to check if
// the number n is prime or not
static bool isPrime(int n)
{
int i;
// Base Cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// Check to skip middle five
// numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
{
return false;
}
for(i = 5; i * i <= n; i += 6)
{
// If n is divisible by i & i+2
// then it is not prime
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}
// Function to find number of subarrays
// with sum exactly equal to k
static int findSubarraySum(int []arr, int n,
int K)
{
// STL map to store number of subarrays
// starting from index zero having
// particular value of sum.
Dictionary prevSum = new Dictionary();
int res = 0;
// To store the sum of element traverse
// so far
int currsum = 0;
for(int i = 0; i < n; i++)
{
// Add current element to currsum
currsum += arr[i];
// If currsum = K, then a new
// subarray is found
if (currsum == K)
{
res++;
}
// If currsum > K then find the
// no. of subarrays with sum
// currsum - K and exclude those
// subarrays
if (prevSum.ContainsKey(currsum - K))
{
res += (prevSum[currsum - K]);
}
// Add currsum to count of
// different values of sum
if (prevSum.ContainsKey(currsum))
{
prevSum[currsum] = prevSum[currsum] + 1;
}
else
{
prevSum.Add(currsum, 1);
}
}
// Return the readonly result
return res;
}
// Function to count the subarray with K primes
static void countSubarray(int []arr, int n, int K)
{
// Update the array element
for(int i = 0; i < n; i++)
{
// If current element is prime
// then update the arr[i] to 1
if (isPrime(arr[i]))
{
arr[i] = 1;
}
// Else change arr[i] to 0
else
{
arr[i] = 0;
}
}
// Function Call
Console.Write(findSubarraySum(arr, n, K));
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4 };
int K = 2;
int N = arr.Length;
// Function Call
countSubarray(arr, N, K);
}
}
// This code is contributed by 29AjayKumar 输出：
4
时间复杂度： O(N * log(log(N)))