Subarray Divisibility
Key Idea: Prefix sums modulo n, counting subarrays with remainder 0 via combinatorics on matching remainders.
Solution
#include <bits/stdc++.h>
using namespace std;
//author: von_Braun
#define ll long long
#define lli long long int
#define pb push_back
#define rep(var, start, num) for(ulli var = start; var <start + num; var++)
#define all(x) x.begin(), x.end()
#define ulli unsigned long long int
#define ull unsigned long long
bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b) { return (a.second < b.second); }
void solve() {
int n;
cin>>n;
vector<ll> a(n);
rep(i,0,n) {cin>>a[i];}
vector<ll> pf(n);
vector<ll> cnts(n,0);
pf[0]=(a[0]%n+n)%n;
cnts[pf[0]]++;
rep(i,1,n-1) {
pf[i] = ((pf[i-1]%n+n)%n + (a[i]%n+n)%n)%n;
cnts[pf[i]]++;
}
ll z=0;
rep(i,0,n) {
ll tp = (cnts[i]*(cnts[i]-1))/2;
z+=tp;
}
z+=cnts[0];
cout<<z<<endl;
}
int main() {
//add quotes incase input output file
//freopen(input.txt,r,stdin);
//freopen(output.txt,w,stdout);
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int tc = 1;
// cin >> tc;
for (int t = 1; t <= tc; t++) {
solve();
}
}