# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
380024 | JerryLiu06 | Zoltan (COCI16_zoltan) | C++11 | 417 ms | 25596 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, ll> pii;
#define pb push_back
#define f first
#define s second
int N, A[200010]; ll powr[100010]; vector<int> allX;
pii incr[800010]; pii decr[800010]; pii DP1[200010], DP2[200010];
const ll MOD = 1000000007LL;
int getX(int X) { return lower_bound(allX.begin(), allX.end(), X) - allX.begin() + 1; }
pii comb(pii A, pii B) { if (A.f != B.f) return max(A, B); return pii {A.f, (A.s + B.s) % MOD}; }
void update(pii tree[], int x, int l, int r, int pos, pii val) { int mid = (l + r) / 2;
if (r < pos || l > pos) return ; if (l == r) { tree[x] = val; return ; }
update(tree, 2 * x, l, mid, pos, val); update(tree, 2 * x + 1, mid + 1, r, pos, val);
tree[x] = comb(tree[2 * x], tree[2 * x + 1]);
}
pii query(pii tree[], int x, int l, int r, int tl, int tr) { int mid = (l + r) / 2;
if (r < tl || l > tr) return pii {0, 0}; if (tl <= l && r <= tr) return tree[x];
return comb(query(tree, 2 * x, l, mid, tl, tr), query(tree, 2 * x + 1, mid + 1, r, tl, tr));
}
int main() {
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> N; for (int i = 0; i < N; i++) { cin >> A[i]; allX.pb(A[i]); }
powr[0] = 1; for (int i = 1; i <= N; i++) powr[i] = (2 * powr[i - 1]) % MOD;
sort(allX.begin(), allX.end()); allX.resize(distance(allX.begin(), unique(allX.begin(), allX.end())));
// cout << MX << endl;
int MX = 0; for (int i = N - 1; i >= 0; i--) { int X = getX(A[i]);
DP1[i] = query(incr, 1, 1, N, X + 1, N); DP1[i].f++; if (DP1[i].s == 0) DP1[i].s = 1;
DP2[i] = query(decr, 1, 1, N, 1, X - 1); DP2[i].f++; if (DP2[i].s == 0) DP2[i].s = 1;
// cout << DP1[i].f << " " << DP2[i].f << endl;
update(incr, 1, 1, N, X, DP1[i]); update(decr, 1, 1, N, X, DP2[i]);
MX = max(MX, DP1[i].f + DP2[i].f - 1); // cout << MX << endl;
}
ll ans = 0; for (int i = 0; i < N; i++) if (DP1[i].f + DP2[i].f - 1 == MX) { ans += DP1[i].s * DP2[i].s; ans %= MOD; }
ans *= powr[N - MX]; ans %= MOD; cout << MX << " " << ans << endl;
}
Compilation message (stderr)
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |