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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#158726#7108. Couleurucup-team045#AC ✓1784ms28392kbC++204.0kb2023-09-02 16:54:552023-09-02 16:54:56

Judging History

你现在查看的是最新测评结果

  • [2023-09-02 16:54:56]
  • 评测
  • 测评结果:AC
  • 用时:1784ms
  • 内存:28392kb
  • [2023-09-02 16:54:55]
  • 提交

answer

#include<iostream>
#include<cstring>
#include<vector>
#include<set>
#include<array>
using namespace std;
using LL = long long;

using Info = int;

const int maxn = 1e5 + 5;
struct Node{
    Info info;
    int l, r;
}tr[maxn * 64];
int root[maxn];
int idx;

void modify_from(int &now, int pre, int l, int r, int x, int v){
    tr[now = ++idx] = tr[pre];
    if (l == r){
        tr[now].info += v;
        return;
    }
    int mid = (l + r) / 2;
    if (x <= mid) modify_from(tr[now].l, tr[pre].l, l, mid, x, v);
    else modify_from(tr[now].r, tr[pre].r, mid + 1, r, x, v);
    tr[now].info = tr[tr[now].l].info + tr[tr[now].r].info;
}

Info query(int u, int l, int r, int L, int R){
    if (r < L || l > R){
        return Info();
    }
	if (l >= L && r <= R){
	 	return tr[u].info;
	}
	int mid = (l + r) / 2;
	return query(tr[u].l, l, mid, L, R) + query(tr[u].r, mid + 1, r, L, R);
}

Info query(int root1, int root2, int l, int r, int L, int R){
    if (r < L || l > R){
        return Info();
    }
    if (l >= L && r <= R){
        return tr[root2].info - tr[root1].info;
    }
    int mid = (l + r) / 2;
    return query(tr[root1].l, tr[root2].l, l, mid, L, R) + query(tr[root1].r, tr[root2].r, mid + 1, r, L, R);
}

struct Seg{
    int l, r;
    LL ans;
    bool operator<(const Seg &t) const{
        if (r != t.r) return r < t.r;
        return l < t.l;
    }
};

int main(){

#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        vector<int> a(n + 1);
        for(int i = 1; i <= n; i++) cin >> a[i];        
        for(int i = 0; i <= idx; i++)
            tr[i] = {0, 0, 0};
        idx = 0;
        for(int i = 1; i <= n; i++) root[i] = 0;
        for(int i = 1; i <= n; i++){
            modify_from(root[i], root[i - 1], 1, n, a[i], 1);
        }
        LL last = 0;
        for(int i = 1; i <= n; i++)
            last += query(root[i - 1], 1, n, a[i] + 1, n);
        set<Seg> s;
        multiset<LL> mx;
        mx.insert(last);
        s.insert({1, n, last});
        for(int i = 1; i <= n; i++){
            cout << last << " \n"[i == n];
            LL x;
            cin >> x;
            if (i == n) break;
            x ^= last;
            auto it = s.lower_bound({0, (int)x, 0});
            auto [l, r, invs] = *it;
            s.erase(it);
            mx.erase(mx.lower_bound(invs));
            invs -= query(root[l - 1], root[x - 1], 1, n, a[x] + 1, n);
            invs -= query(root[x], root[r], 1, n, 1, a[x] - 1);
            if (x - l <= r - x){
                LL invs1 = 0, invs2 = 0;
                for(int i = l; i < x; i++){
                    invs1 += query(root[i], root[x - 1], 1, n, 1, a[i] - 1);
                    invs2 += query(root[x], root[r], 1, n, 1, a[i] - 1);
                }
                invs2 = invs - invs2 - invs1;
                if (l <= x - 1){
                    s.insert({l, (int)x - 1, invs1});
                    mx.insert(invs1);
                }
                if (x + 1 <= r){
                    s.insert({(int)x + 1, r, invs2});
                    mx.insert(invs2);
                }
            }
            else{
                LL invs1 = 0, invs2 = 0;
                for(int i = x + 1; i <= r; i++){
                    invs2 += query(root[x], root[i - 1], 1, n, a[i] + 1, n);
                    invs1 += query(root[l - 1], root[x - 1], 1, n, a[i] + 1, n);
                }
                invs1 = invs - invs1 - invs2;
                if (l <= x - 1){
                    s.insert({l, (int)x - 1, invs1});
                    mx.insert(invs1);
                }
                if (x + 1 <= r){
                    s.insert({(int)x + 1, r, invs2});
                    mx.insert(invs2);
                }
            }
            last = *mx.rbegin();
        }
    }

}

这程序好像有点Bug,我给组数据试试?

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 5700kb

input:

3
5
4 3 1 1 1
5 4 5 3 1
10
9 7 1 4 7 8 5 7 4 8
21 8 15 5 9 2 4 5 10 6
15
4 8 8 1 12 1 10 14 7 14 2 9 13 10 3
37 19 23 15 7 2 10 15 2 13 4 5 8 7 10

output:

7 0 0 0 0
20 11 7 2 0 0 0 0 0 0
42 31 21 14 14 4 1 1 1 0 0 0 0 0 0

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 1784ms
memory: 28392kb

input:

11116
10
10 5 10 3 6 4 8 5 9 8
31 27 24 11 12 3 0 2 3 1
10
8 2 7 2 8 10 1 10 9 10
6 5 2 13 2 1 0 1 3 1
10
7 10 7 6 1 3 10 6 7 9
21 18 10 1 6 5 4 8 9 10
10
2 10 4 8 8 5 7 2 6 7
20 10 9 1 15 0 4 2 9 7
10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
10
1 2 3 4 5 6 7 8 9 10
6 3 5 2 7 10 9 1 4 8
10
1 10 1 3...

output:

21 18 16 12 10 6 4 1 1 0
12 12 10 10 4 4 4 2 1 0
20 16 9 5 3 3 3 0 0 0
22 14 8 8 5 5 2 1 1 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
19 12 7 4 4 2 2 1 0 0
20 18 8 3 1 1 0 0 0 0
45 21 21 10 3 3 3 0 0 0
17 11 8 2 1 1 1 0 0 0
13 4 1 0 0 0 0 0 0 0
29 27 22 15 9 7 4 3 1 0
26 16 9 2 1 1 1 1 1 0
0 0 0 0 0 ...

result:

ok 11116 lines

Extra Test:

score: 0
Extra Test Passed