Given an array find the number of sub arrays with m odd numbers?

Sample Input: 1 2 3 4 5 (array elements) m = 1 (Odd numbers) Sample output: 8. The subarrays are [[1], [1,2], [2,3], [2,3,4], [3], [3,4], [4,5], [5] ]

Here is my implementation.In the worst case, it would take O(n+n^2).Are there any ways to optimize this code?

int main() {
    int n, *a, count1=0, m, *dp;
    cin>>n;
    a = new int[n];
    dp =new int[n];

    for(int i=0; i<n; i++) {
        cin >> a[i];
    }

    cin >> m;

    for(int i=0; i<n; i++){
        if(a[i]%2==1) {
            count1++;
        }

        dp[i] =count1;
    }

    int prev;
    long count=0;

    for(int i=0; i<n; i++) {
        prev= i==0 ? 0:dp[i-1];
        for(int j=i; j<n; j++) {
            if((dp[j] - prev) == m){
                count++;
            }

            if(dp[j] - prev > m){
                break;
            }
        }
    }
    cout<<count;
}