Given an integer pair \((l,r)\) \((1\leq l\leq r\leq N)\), consider how we can determine if the condition is held.
For positive integers \(X\) and \(Y\), the conditions “\(X\) is a divisor of \(Y\)” and “the remainder when dividing \(Y\) by \(X\) is \(0\)” are equivalent. Therefore, the steps required are:

• find \(S=A_l+A_{l+1}+\cdots+A_r\), and
• for each \(l\leq i\leq r\), find the remainder when dividing \(S\) by \(A_i\). If any \(i\) yields a remainder \(0\), then \((l,r)\) does not satisfy the condition; otherwise, it does.

The two steps can be performed with for statements and basic operations. The computational complexity is \(O(N)\).

The number of candidate pairs \((l,r)\) is \(\frac{N(N+1)}{2}\), and they can be enumerated with for statements. For each of them we check the condition above, and count the number of applicable pairs. The time complexity is \(O(N^3)\), which is fast enough under the constraints of the problem.

Sample code in C++:

#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	int a[50]={};
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}

	int ans=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<=i;j++){
			int s=0;
			for(int k=j;k<=i;k++)s+=a[k];
			bool ok=true;
			for(int k=j;k<=i;k++)if(s%a[k]==0)ok=false;
			if(ok)ans++;
		}
	}
	cout<<ans<<endl;
	return 0;
}

Sample code in Python:

n=int(input())
a=list(map(int, input().split()))
ans=0

for i in range(n):
    for j in range(i+1):
        s=0
        for k in range(j,i+1):
            s+=a[k]
        flag=True
        for k in range(j,i+1):
            if s%a[k]==0:
                flag=False
        if flag:
            ans+=1

print(ans)