Submission #1203134

#	Time	Username	Problem	Language	Result	Execution time	Memory
1203134		SpyrosAliv	Arranging Shoes (IOI19_shoes)	C++20	45 / 100	50 ms	6948 KiB

shoes

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

int n;
vector<int> a;

class Segtree {
    private:
    int n;
    vector<int> seg;
    void update(int node, int l, int r, int idx) {
        if (l == r) seg[node] = 1;
        else {
            int mid = (l + r) / 2;
            if (idx <= mid) update(node*2+1, l, mid, idx);
            else update(node*2+2, mid+1, r, idx);
            seg[node] = seg[node*2 + 1] + seg[node*2 + 2]; 
        }
    }
    int query(int node, int start, int end, int l, int r) {
        if (start > r || end < l) return 0;
        else if (start >= l && end <= r) return seg[node];
        else {
            int mid = (start + end) / 2;
            return query(node*2+1, start, mid, l, r) + query(node*2+2, mid+1, end, l, r);
        }
    }
    public:
    Segtree(int nn) {
        n = nn;
        seg.assign(n*4, 0);
    }
    void update(int idx) {
        update(0, 0, n-1, idx);
    }
    int query(int l, int r) {
        if (l > r) return 0;
        return query(0, 0, n-1, l, r);
    }
};

ll count_swaps(vector<int> s) {
    n = s.size();
    a = s;
    Segtree seg(n);
    vector<int> l, r;
    for (int i = 0; i < n; i++) {
        if (s[i] < 0) l.push_back(i);
        else r.push_back(i);
    }
    reverse(l.begin(), l.end());
    reverse(r.begin(), r.end());
    ll ans = 0;
    for (int i = 0; i < n; i++) {
        int curr = 0;
        if (i % 2 == 0) {
            curr = l.back();
            l.pop_back();
        }
        else {
            curr = r.back();
            r.pop_back();
        }
        int nxt = curr + seg.query(curr+1, n-1);
        ans += abs(nxt - i);
        seg.update(curr);
    }
    return ans;
}
/*
int main() {
    int n; cin >> n;
    vector<int> arr;
    for (int i = 0; i < n; i++) {
        int x; cin >> x;
        arr.push_back(x);
    }
    cout << count_swaps(arr) << "\n";
}*/

• Subtask 1: 10 / 10
• Subtask 2: 0 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 15