使用C++編寫，找到給定範圍內前綴和質數的數量

Input : arr[ ] = { 3, 5, 6, 2, 4 }
L = 1, R = 3
Output : 3
Explanation : prefixsum[ 0 ] = arr[ L ] = 5
prefixsum[ 1 ] = prefixsum[ 0 ] + arr[ 2 ] = 11
prefixsum[ 2 ] = prefixsum[ 1 ] + arr[ 3 ] = 13
In prefixsum[ ] array all three 5, 11 and 13 are prime numbers in prefix sum array in given range.

Input : arr[ ] = { 6, 10, 5, 8, 11 }
L = 0, R = 3
Output : 1
Explanation : prefixsum[ 0 ] = arr[ L ] = 6
prefixsum[ 1 ] = prefixsum[ 0 ] + arr[ 1 ] = 16
prefixsum[ 2 ] = prefixsum[ 1 ] + arr[ 2 ] = 21
prefixsum[ 3 ] = prefixsum[ 2 ] + arr[ 3 ] = 29
In prefixsum[ ] array only 29 is the prime number in prefix sum array given range.

#include<bits/stdc++.h>
using namespace std;
vector < bool > checkprime (int *arr, int n, int MAX){
    vector < bool > p (n);
    bool Prime_val[MAX + 1];
    for (int i = 2; i < MAX; i++)
        Prime_val[i] = true;
        Prime_val[1] = false;
    for (int p = 2; p * p <= MAX; p++){
        // If prime[p] is not changed, then
        // it is a prime
        if (Prime_val[p] == true){
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                Prime_val[i] = false;
        }
    }
    for (int i = 0; i < n; i++){
        if (Prime_val[arr[i]])
             p[i] = true;
        else
        p[i] = false;
    }
    return p;
}
int main (){
    int arr[] = { 2, 3, 4, 7, 9, 10 };
    int s1 = sizeof (arr) / sizeof (arr[0]);//size of given array
    int L = 1, R = 3, s2 = R - L + 1;
    int prefixsum[s2];
    int count = 0;
    prefixsum[0] = arr[L];
    for (int i = L + 1, j = 1; i <= R && j < s1; i++, j++){
        prefixsum[j] = prefixsum[j - 1] + arr[i];

    }
    vector < bool > isprime = checkprime (prefixsum, s2, prefixsum[s2 - 1]);
    for (int i = 0; i < s2; i++) {
        if (isprime[i] == 1)
            count++;
    }
    cout <<"Number of prefix sum prime in given range query: " << count;
    return 0;
}

Number of prefix sum prime in given range query: 2