Submission #1262918

#	Time	Username	Problem	Language	Result	Execution time	Memory
1262918		trinhvtuan	Sum Zero (RMI20_sumzero)	C++20	61 / 100	322 ms	36708 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 (value, pos) including pos 0
    vector<pair<ll,int>> vec;
    vec.reserve(n+1);
    ll cur = 0;
    vec.emplace_back(0LL, 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 (0 <= curpos && curpos <= n) lf[curpos] = prevpos;
        }
    }
    // accumulate so lf[i] = best available previous occurrence up to i
    for(int i=1;i<=n;i++) if (lf[i] < lf[i-1]) lf[i] = lf[i-1];

    // free vec to reduce peak memory
    vec.clear(); vec.shrink_to_fit();

    // compute LOG (number of layers for binary lifting)
    int LOG = 1;
    while((1<<LOG) <= n) ++LOG; // now 2^LOG > n
    // use 1D array to store st: st[k*(n+1) + i]
    int rows = LOG;
    int cols = n + 1;
    vector<int> st(rows * cols, -1);

    auto idx = [&](int k, int i){ return k * cols + i; };

    // initialize level 0 for all positions 0..n
    for(int i=0;i<=n;i++){
        st[idx(0,i)] = (i>=0 && i<=n) ? lf[i] : -1; // lf[0] is -1 normally
    }
    // build higher levels
    for(int k=1;k<rows;k++){
        for(int i=0;i<=n;i++){
            int mid = st[idx(k-1,i)];
            st[idx(k,i)] = (mid == -1) ? -1 : st[idx(k-1, mid)];
        }
    }

    int Q; cin >> Q;
    while(Q--){
        int L,R; cin >> L >> R;
        int u = R;
        int ans = 0;
        for(int k=rows-1;k>=0;k--){
            int to = st[idx(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