Submission #121438

#	Time	Username	Problem	Language	Result	Execution time	Memory
121438		RockyB	Lottery (CEOI18_lot)	C++17	0 / 100	846 ms	8856 KiB

lot

/// In The Name Of God

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>

#define f first
#define s second

#define pb push_back
#define pp pop_back
#define mp make_pair

#define sz(x) (int)x.size()
#define sqr(x) ((x) * 1ll * (x))
#define all(x) x.begin(), x.end()

#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, l, r) for (int i = (l); i >= (r); i--)

#define Kazakhstan ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

#define nl '\n'
#define ioi exit(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = (int)1e4 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;

const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};

using namespace std;

int n, q, l;
int a[N];

int cnt1[N], cnt2[N];
pair <int, int> k[N];
int pref[N][102], ans[N][102];
int main() {
	#ifdef IOI2018
		freopen ("in.txt", "r", stdin);
		//freopen ("slow.out", "w", stdout);
	#endif
	Kazakhstan
	cin >> n >> l;
	rep(i, 1, n) cin >> a[i];
	cin >> q;	
	rep(i, 1, q) {
		cin >> k[i].f;
		k[i].s = i;
	}
	sort (k + 1, k + 1 + q);
	rep(d, 0, n - l) {
		int l1 = 1, l2 = 1 + d, df = 0;
		rep(i, 1, n - l + 1) {
			if (a[i] != a[i + d]) ++df;
			while (i - l1 + 1 > l) {
				if (a[l1] != a[l2]) --df;
				l1++, l2++;
			}
			if (i >= l) {
				cnt1[l1] = df;
				cnt2[l2] = df;
			}
		}
		rep(i, 1, n) {
			if (i + d + l - 1 <= n) { 
				int pos = lower_bound (k + 1, k + 1 + q, mp(cnt1[i], -inf)) - k;
				//cout << i << ' ' << cnt1[i] << ' ' << pos << nl;
				pref[i][pos]++;
			}
			if (i + l - 1 <= n && i - d > 0 && d) {
				int pos = lower_bound (k + 1, k + 1 + q, mp(cnt2[i], -inf)) - k;
				pref[i][pos]++;
			}
		}
		cout << nl;
	}
	rep(i, 1, n) {
		rep(j, 1, q) {
			pref[i][j] += pref[i][j - 1];
			ans[i][k[j].s] = pref[i][j];
		}
	}
	rep(i, 1, q) {
		rep(j, 1, n - l + 1) {
			cout << ans[j][i] - 1 << ' ';
		}
		cout << nl;
	}
	ioi
}

• Subtask 1: 0 / 25
• Subtask 2: 0 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20