(Analysis by Nick Wu)

Because the string length can be as large as $10^5$, writing three for loops to manually count every instance of the word COW will be too slow. We won't even be able to have two nested for loops.

Let's imagine looping only once, going across the string from left-to-right. There are three types of sequences that are important to us - the number of times the sequence "C" appears, the number of times the sequence "CO" appears, and the number of times the sequence "COW" appears.

If the next character is a "W", then the number of times "COW" appears increases by the number of instances of the sequence "CO" seen so far.

If the next character is a "O", then the number of times "CO" appears increases by the number of instances of the sequence "C" seen so far.

If the next character is a "C", then the number of times "C" appears increases by 1.

At the end of this process, we simply report how many instances of "COW" we have seen.

Here is Mark Gordon's code illustrating this process:

#include <iostream>
#include <vector>

using namespace std;

int main() {
freopen("cow.in", "r", stdin);
freopen("cow.out", "w", stdout);

int N;
string S;
cin >> N >> S;

long long NC = 0;
long long NO = 0;
long long NW = 0;
for (int i = 0; i < N; i++) {
if (S[i] == 'C') {
NC++;
} else if (S[i] == 'O') {
NO += NC;
} else if (S[i] == 'W') {
NW += NO;
}
}

cout << NW << endl;
return 0;
}