Frequency of each element of an array considering all contiguos subarrays

Question

Consider an array A = [5,1,7,2,3]

All contiguos subarrays = { [5], [1], [7], [2], [3], [5,1], [1,7], [7,2], [2,3], [5,1,7], [1,7,2], [7,2,3], [5,1,7,2], [1,7,2,3],

Answer 1

I am having hard time trying to explain my solution in words. I will just add the code. It will explain itself:

#include 
#include 
using namespace std;

#define max 10000

int main(int argc, const char * argv[]) {

    ifstream input("/Users/appleuser/Documents/Developer/xcode projects/SubArrayCount/SubArrayCount/input.in");

    int n, arr[max], before[max]={0}, after[max]={0}, result[max];
    input >> n;
    for (int i=0; i> arr[i];

    for (int i=0;i=0&&arr[j]=0;i--)
        for (int j=i+1;j

Explanation for (before[i]+1)*(after[i]+1):

for each value we need the numbers lies before and less than the value and the numbers lies after and less than the value.

  | 0  1  2  3  4  5 .... count of numbers less than the value and appears before.
---------------------
0 | 1  2  3  4  5  6
1 | 2  4  6  8  10 12
2 | 3  6  9  12 15 18
3 | 4  8  12 16 20 24
4 | 5  10 15 20 25 30
5 | 6  12 18 24 30 36
. | 
. |
. |
count of numbers less than the value and appears after.

Example: for a number that have 3 values less than it and appears before and have 4 values less than it and appears after. answer is V(3,4) = 20 = (3+1) * (4+1)

please, let me know the results.