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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#291198#7685. Barkley IIOAleksaRE 0ms32980kbC++141.5kb2023-12-26 05:31:502023-12-26 05:31:51

Judging History

你现在查看的是最新测评结果

  • [2023-12-26 05:31:51]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:32980kb
  • [2023-12-26 05:31:50]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define f first
#define s second
#define int long long
const int N = 5e5 + 69;
int n, m, a[N], fenw[N];
vector<int> pos[N];
map<int, int> lst, sv;
vector<pair<int, int>> q[N];
signed main() {
	ios_base::sync_with_stdio(false);
  cout.tie(nullptr); 
  cin.tie(nullptr);
  //freopen("newbarn.in", "r", stdin);
  //freopen(newbarn.out", "w", stdout);
  int tt = 1;
  cin >> tt;
  while (tt--) {
  	auto add = [&](int v, int val) {
  		for (int i = v;i < N;i += (i & -i))
  			fenw[i] += val;
  	};
  	auto get = [&](int v) {
  		int res = 0;
  		for (int i = v;i >= 1;i -= (i & -i))
  			res += fenw[i];
  		return res;
  	};
  	cin >> n >> m;
  	for (auto i : lst)
			add(i.f, -1);
		for (auto i : sv)
			q[i.f].clear();
  	sv.clear();
  	lst.clear();
  	for (int i = 1;i <= n;i++) {
  		cin >> a[i];
  		sv[a[i]] = 1;
  	}
  	for (auto u : sv)
  		pos[u.f].push_back(0);
  	for (int i = 1;i <= n;i++)
  		pos[a[i]].push_back(i);
  	for (auto u : sv)
  		pos[u.f].push_back(n + 1);
  	for (auto i : sv) {
  		for (int j = 1;j < (int)pos[i.f].size();j++) {
  			q[pos[i.f][j] - 1].push_back({pos[i.f][j - 1] + 1, i.f});
  		}
  	}

  	int ans = -1;
  	for (int i = 1;i <= n;i++) {
  		if (lst[a[i]] > 0)
  			add(lst[a[i]], -1);
  		add(i, 1);
  		for (auto u : q[i]) {
  			int l = u.f, mx = u.s;
  			ans = max(ans, get(i) - get(l - 1) - mx);
  		}
  		lst[a[i]] = i;
  	}

  	cout << ans << '\n';
  }
  return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 32980kb

input:

2
5 4
1 2 2 3 4
5 10000
5 2 3 4 1

output:

2
3

result:

ok 2 number(s): "2 3"

Test #2:

score: -100
Runtime Error

input:

50000
10 19
12 6 1 12 11 15 4 1 13 18
10 8
8 7 6 7 6 2 2 3 4 8
10 6
3 2 6 6 5 2 3 4 5 6
10 11
6 3 7 9 2 1 2 10 10 4
10 6
6 1 2 6 1 1 3 4 2 1
10 9
8 5 3 9 1 7 5 5 1 1
10 5
1 4 3 2 5 4 5 3 5 2
10 14
3 8 12 10 4 2 3 13 7 3
10 14
5 5 12 2 8 1 13 9 8 5
10 7
5 5 6 6 1 5 3 7 3 4
10 7
5 1 4 6 1 6 4 3 7 5
10...

output:


result: