## § Subarrays ~= prefixes

To solve any problem about subarrays, we can reinterpret a subarray [l..r] as a prefix [0..r] - [0..l]. For example, to find all subarrays [l..r] whose sum of elements divides n, we can think of this as finding a subarray [l..r] where the sum of elements modulo n is zero. This is CSES' subarray divisibiity problem:

int main() {
int n;
cin >> n;
vector<ll> xs(n);
for (int i = 0; i < n; ++i) {
cin >> xs[i]; xs[i] = xs[i] % n; if (xs[i] < 0) { xs[i] += n; }
}

ll count = 0; // number of subarrays with sum = 0 (mod n)
ll cursum = 0; //  current sum [0..i]
// number of subarrays [0..r] (for some r) such that Σa[i] = count.
map<ll, ll> partial_sum_count;
partial_sum_count = 1;

for (int i = 0; i < n; ++i) {
// current sum [0..i]
cursum = (cursum + xs[i]) % n;

// for each [0..j] (for j < i) with sum cursum, we want:
// sum([i..j]) = 0
// => sum([0..i]) - sum([0..j)) = 0
// => sum([0..i]) = sum([0..j))
// for each such j, we get one subarray.
auto it = partial_sum_count.find(cursum);
if (it != partial_sum_count.end()) {
count += it->second;
}

// partial sum [0..i] = cursum
partial_sum_count[cursum]++;
}

cout << count << "\n";

return 0;
}