给定范围查询中前缀和素数的数量

给定一个非负整数数组和范围查询 l, r，找到给定范围内的素数前缀和的数量。
先决条件：前缀总和 |素性检验

例子 ：

Input : {2, 3, 4, 7, 9, 10}, 
        l = 1, r = 5;
Output : 3
Explanation : prefixSum[0] = arr[l] = 3
         prefixSum[1] = prefixSum[0] + arr[2] = 7, 
         prefixSum[2] = prefixSum[1] + arr[3] = 14, 
         prefixSum[3] = prefixSum[2] + arr[4] = 23, 
         prefixSum[4] = prefixSum[3] + arr[5] = 33,
There are three primes in prefix sum array in given
range. The primes are 3, 7 and 23.

Input : {5, 7, 8, 10, 13}, 
         l = 0, r = 4;
Output : 2
         prefixSum[0] = arr[l] = 5, 
         prefixSum[1] = prefixSum[0] + arr[1] = 12, 
         prefixSum[2] = prefixSum[1] + arr[2] = 20, 
         prefixSum[3] = prefixSum[2] + arr[3] = 30, 
         prefixSum[4] = prefixSum[3] + arr[4] = 43, 
There are two primes in prefix sum array in given
range. The primes are 5 and 43

方法：通过 l 到 r 运行一个循环，其中 l 和 r 是索引的范围，并通过添加前缀和数组的前一个元素和当前元素来填充大小与数组相同的前缀和数组数组，然后通过检查素数的Primality Test检查每个阶段的前缀和是否为素数。如果前缀和是素数，则增加计数，否则不增加。

C++

// C++ program to count the number of
// prefix sum which are prime or not
#include
using namespace std;
 
// Primality test to check if prefix
// sum is prime or not
bool isPrime(int num)
{
    // Corner case
    if (num <= 1) return false;
    if (num <= 3) return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (num%2 == 0 || num%3 == 0)
        return false;
 
    for (int j = 5; j * j <= num; j = j + 6)
        if (num%j == 0 || num%(j+2) == 0)
        return false;
 
    return true;
}
 
// to calculate the prefix sum for
// the given range of query
int primesubArraySum(int arr[], int n, int l,
                     int r, int preSum[])
{
    int count = 0;   
    preSum[0] = arr[l];   
    if (isPrime(preSum[0]))
       count++;
     
     
    for (int i = l + 1, j = 1;
        i <= r && j < n; i++,j++)
    {
        preSum[j] = preSum[j - 1] + arr[i];
         
        // increase the count if the
        // prefix sum is prime
        if (isPrime(preSum[j]))
            count++;
    }
     
    return count;
}
 
//driver code
int main()
{
    int arr[] = {5, 7, 8, 10, 13};
    int n = sizeof(arr)/sizeof(arr[0]);
    int preSum[n];
    int l = 0, r = 4;
    
    cout << primesubArraySum(arr, n, l, r, preSum);
     
    return 0;
}

Java

// Java program to count the number of
// prefix sum which are prime or not
import java.util.*;
class GFG
{
    // Primality test to check if
    // prefix sum is prime or not
    public static boolean isPrime(int num)
    {
        // Corner cases
        if (num <= 1)
            return false;
        if (num <= 3)
            return true;
     
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (num % 2 == 0 || num % 3 == 0)
                return false;
     
        for (int j = 5; j * j <= num; j = j + 6)
            if (num % j == 0 || num % (j + 2) == 0)
                return false;
     
        return true;
    }
     
    // to calculate the prefix sum for
    // the given range of query
    public static int primesubArraySum(int arr[], int n,
                                       int l, int r,
                                       int preSum[])
    {
        int count = 0;
     
        preSum[0] = arr[l];
     
        if (isPrime(preSum[0]) == true)
            count++;
     
        for (int i = l+1,j = 1; i <= r && j < n; i++,j++)
        {
            preSum[j] = preSum[j-1] + arr[i];
         
            // increase the count if the prefix sum is prime
            if (isPrime(preSum[j]) == true)
                count++;
        }
     
        return count;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = {5, 7, 8, 10, 13};
        int n = arr.length;
        int preSum[] = new int[n];
        int l = 0, r = 4;
        System.out.println(primesubArraySum(arr, n,
                           l, r, preSum));
    }
}

Python3

# Python program to count the number
# of prefix sum which are prime or not
import numpy
import math
 
# Primality test to check if prefix
# sum is prime or not
def isPrime(num):
     
    # Corner case
    if (num <= 1):
        return False;
    if (num <= 3):
        return True;
 
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (num % 2 == 0 or num % 3 == 0):
        return False;
 
    for j in range(5, (int)(math.sqrt(num)) + 1, 6):
        if (num % j == 0 or num % (j + 2) == 0):
            return False;
 
        return True;
 
# to calculate the prefix sum for
# the given range of query
def primesubArraySum(arr, n, l, r, preSum):
 
    count = 0;
    preSum[0] = arr[l];
 
    if (isPrime(preSum[0])):
        count = count + 1;
     
     
    for i in range(l + 1, r):
        for j in range(1, n):
            preSum[j] = preSum[j - 1] + arr[i];
             
            # increase the count if the
            # prefix sum is prime
            if (isPrime(preSum[j])):
                count = count + 1;
 
    return count;
 
# Driver code
arr = [5, 7, 8, 10, 13];
n = len(arr);
 
#a = numpy.arange(5)
preSum = numpy.arange(n);
l = 0;
r = 4;
print(primesubArraySum(arr, n, l, r, preSum));
     
# This code is contributed
# by Shivi_Aggarwal

C#

// C# program to count the number of
// prefix sum which are prime or not
using System;
 
class GFG {
     
    // Primality test to check if
    // prefix sum is prime or not
    public static bool isPrime(int num)
    {
         
        // Corner cases
        if (num <= 1)
            return false;
        if (num <= 3)
            return true;
     
        // This is checked so that we can skip
        // middle five numbers in below loop
        if (num % 2 == 0 || num % 3 == 0)
                return false;
     
        for (int j = 5; j * j <= num; j = j + 6)
            if (num % j == 0 || num % (j + 2) == 0)
                return false;
     
        return true;
    }
     
    // To calculate the prefix sum
    // for the given range of query
    public static int primesubArraySum(int []arr, int n,
                                       int l, int r,
                                       int []preSum)
    {
        int count = 0;
     
        preSum[0] = arr[l];
     
        if (isPrime(preSum[0]) == true)
            count++;
     
        for (int i = l+1,j = 1; i <= r &&
                          j < n; i++,j++)
        {
            preSum[j] = preSum[j-1] + arr[i];
         
            // increase the count if the
            // prefix sum is prime
            if (isPrime(preSum[j]) == true)
                count++;
        }
     
        return count;
    }
 
    // Driver code
    public static void Main ()
    {
        int []arr = {5, 7, 8, 10, 13};
        int n = arr.Length;
        int []preSum = new int[n];
        int l = 0, r = 4;
        Console.Write(primesubArraySum(arr, n,
                      l, r, preSum));
    }
}
 
// This code is contributed by Nitin Mittal.

Javascript

 
// Javascript program to count the number of
// prefix sum which are prime or not
 
// Primality test to check if
// prefix sum is prime or not
function isPrime(num)
{
     
    // Corner cases
    if (num <= 1)
        return false;
    if (num <= 3)
        return true;
   
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (num % 2 == 0 || num % 3 == 0)
            return false;
   
    for(let j = 5; j * j <= num; j = j + 6)
        if (num % j == 0 || num % (j + 2) == 0)
            return false;
   
    return true;
}
   
// To calculate the prefix sum
// for the given range of query
function primesubArraySum(arr, n, l, r, preSum)
{
    let count = 0;
   
    preSum[0] = arr[l];
   
    if (isPrime(preSum[0]) == true)
        count++;
   
    for(let i = l + 1, j = 1;
            i <= r && j < n;
            i++, j++)
    {
        preSum[j] = preSum[j - 1] + arr[i];
       
        // Increase the count if the
        // prefix sum is prime
        if (isPrime(preSum[j]) == true)
            count++;
    }
    return count;
}
 
// Driver code
let arr = [ 5, 7, 8, 10, 13 ];
let n = arr.length;
let preSum = new Array(n);
 
preSum.fill(0);
let l = 0, r = 4;
 
document.write(primesubArraySum(
    arr, n, l, r, preSum));
 
// This code is contributed by decode2207
 

输出：

2