Submission #1262917

#	Time	Username	Problem	Language	Result	Execution time	Memory
1262917		trinhvtuan	Sum Zero (RMI20_sumzero)	C++20	61 / 100	419 ms	38340 KiB

sumzero

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

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    if(!(cin >> n)) return 0;
    vector<ll> a(n+1);
    for(int i=1;i<=n;i++) cin >> a[i];

    // build prefix sums and pairs (value, pos)
    vector<pair<ll,int>> vec;
    vec.reserve(n+1);
    ll cur = 0;
    vec.emplace_back(0LL, 0); // prefix sum at position 0
    for(int i=1;i<=n;i++){
        cur += a[i];
        vec.emplace_back(cur, i);
    }

    sort(vec.begin(), vec.end(), [](const pair<ll,int>& A, const pair<ll,int>& B){
        if (A.first != B.first) return A.first < B.first;
        return A.second < B.second;
    });

    // lf[pos] = previous position with same prefix sum (or -1)
    vector<int> lf(n+1, -1);
    for(size_t i=1;i<vec.size();++i){
        if (vec[i].first == vec[i-1].first){
            int curpos = vec[i].second;
            int prevpos = vec[i-1].second;
            if (curpos >= 0 && curpos <= n) lf[curpos] = prevpos;
        }
    }

    // We don't need vec anymore -> free memory
    vec.clear();
    vec.shrink_to_fit();

    // accumulate so lf[i] is the best (rightmost) previous occurrence up to i
    for(int i=1;i<=n;i++){
        if (lf[i] < lf[i-1]) lf[i] = lf[i-1];
    }

    // binary lifting table
    int LOG = 1;
    while((1<<LOG) <= n) ++LOG;
    // st[k][i] = position after jumping 2^k steps from i (or -1)
    vector<vector<int>> st(LOG, vector<int>(n+1, -1));
    for(int i=1;i<=n;i++) st[0][i] = lf[i];

    for(int k=1;k<LOG;k++){
        for(int i=1;i<=n;i++){
            int mid = st[k-1][i];
            st[k][i] = (mid == -1) ? -1 : st[k-1][mid];
        }
    }

    int Q; cin >> Q;
    while(Q--){
        int L,R; cin >> L >> R;
        int u = R;
        int ans = 0;
        for(int k=LOG-1;k>=0;k--){
            int to = st[k][u];
            if (to != -1 && to + 1 >= L){
                ans += (1<<k);
                u = to;
            }
        }
        cout << ans << '\n';
    }
    return 0;
}

• Subtask 1: 22 / 22
• Subtask 2: 39 / 39
• Subtask 3: 0 / 39