Submission #121467

#TimeUsernameProblemLanguageResultExecution timeMemory
121467RockyBLottery (CEOI18_lot)C++17
100 / 100
1296 ms13104 KiB
/// 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)2e4 + 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][110], ans[N][110]; int dp[N]; inline int get(int val) { if (~dp[val]) return dp[val]; int pos = lower_bound (k + 1, k + 1 + q, mp(val, -inf)) - k; //if (pos > q + 1) ioi return dp[val] = pos; } 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); memset(dp, -1, sizeof(dp)); rep(d, 1, n - l) { int l1 = 1, l2 = 1 + d, df = 0; rep(i, 1, n) { 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 = get(cnt1[i]); //cerr << i << ' ' << cnt1[i] << ' ' << pos << nl; pref[i][pos]++; } if (i + l - 1 <= n && i - d > 0) { int pos = get(cnt2[i]); 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] << ' '; } cout << nl; } ioi }
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