Thank you so much for participating in our round! We really hope you enjoyed it. For us, it was an amazing and educational experience, and we look forward to the possibility that we could one day help to or hold a round again. Thanks again to abc864197532 and Vladithur for making the process so smooth and enjoyable.
2027A - Расстановка прямоугольников
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int maxw = 0, maxh = 0;
for (int i=0; i<n; i++) {
int w, h;
cin >> w >> h;
maxw = max(maxw, w);
maxh = max(maxh, h);
}
cout << 2 * (maxw + maxh) << "\n";
}
return 0;
}
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> A(n);
for (int i=0; i<n; i++) cin >> A[i];
int best = 0;
for (int i=0; i<n; i++) {
int curr = 0;
for (int j=i; j<n; j++) {
if (A[j] <= A[i]) {
curr += 1;
}
}
best = max(best, curr);
}
cout << n - best << "\n";
}
return 0;
}
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<ll> A(n);
for (int i=0; i<n; i++) cin >> A[i];
map<ll,vector<ll>> adj;
for (int i=1; i<n; i++) {
ll u = A[i] + i;
ll v = u + i;
adj[u].push_back(v);
}
set<ll> vis;
function<void(ll)> dfs = [&](ll u) -> void {
if (vis.count(u)) return;
vis.insert(u);
for (ll v : adj[u]) dfs(v);
};
dfs(n);
cout << *vis.rbegin() << "\n";
}
return 0;
}
2027D1 - Финальный диалог (простая версия)
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int inf = 1 << 30;
void chmin(int &a, int b) {
a = min(a, b);
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
vector<int> A(n);
for (int i=0; i<n; i++) cin >> A[i];
vector<int> B(m);
for (int i=0; i<m; i++) cin >> B[i];
vector nxt(n, vector<int>(m));
for (int k=0; k<m; k++) {
int r = -1, sum = 0;
for (int i=0; i<n; i++) {
while (r < n && sum <= B[k]) sum += A[++r];
nxt[i][k] = r;
sum -= A[i];
}
}
vector dp(n+1, vector<int>(m, inf));
dp[0][0] = 0;
for (int k=0; k<m; k++) {
for (int i=0; i<n; i++) {
chmin(dp[nxt[i][k]][k], dp[i][k] + m - k - 1);
if (k < m-1)
chmin(dp[i][k+1], dp[i][k]);
}
}
int ans = inf;
for (int k=0; k<m; k++) {
chmin(ans, dp[n][k]);
}
if (ans == inf) {
cout << "-1\n";
} else {
cout << ans << "\n";
}
}
return 0;
}
2027D2 - Финальный диалог (сложная версия)
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int inf = 1 << 30;
const int MOD = 1000000007;
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
vector<int> A(n);
for (int i=0; i<n; i++) cin >> A[i];
vector<int> B(m);
for (int i=0; i<m; i++) cin >> B[i];
vector nxt(n, vector<int>(m));
for (int k=0; k<m; k++) {
int r = -1, sum = 0;
for (int i=0; i<n; i++) {
while (r < n && sum <= B[k]) sum += A[++r];
nxt[i][k] = r;
sum -= A[i];
}
}
vector dp(n+1, vector<array<int,2>>(m, {inf, 0}));
vector upd(n+1, vector(m, vector<array<int,3>>()));
upd[0][0].push_back({0, 0, 1});
upd[1][0].push_back({1, 0, 1});
for (int k=0; k<m; k++) {
map<int,array<int,2>> mp;
for (int i=0; i<=n; i++) {
for (auto [t, move, count] : upd[i][k]) {
if (t == 0) {
auto &[a, b] = mp[move];
a += 1;
(b += count) %= MOD;
} else {
auto &[a, b] = mp[move];
a -= 1;
(b += MOD - count) %= MOD;
if (a == 0) mp.erase(move);
}
}
if (mp.empty()) continue;
auto &[move, info] = *mp.begin();
dp[i][k] = {move, info[1]};
if (i == n) continue;
if (k < m-1) {
upd[i][k+1].push_back({0, move, info[1]});
upd[i+1][k+1].push_back({1, move, info[1]});
}
if (nxt[i][k] > i) {
upd[i+1][k].push_back({0, move + (m - k - 1), info[1]});
if (nxt[i][k] < n) {
upd[nxt[i][k]+1][k].push_back({1, move + (m - k - 1), info[1]});
}
}
}
}
map<int,int> mp;
for (int k=0; k<m; k++) {
auto &[move, count] = dp[n][k];
(mp[move] += count) %= MOD;
}
auto &[move, count] = *mp.begin();
if (move == inf) {
cout << "-1\n";
} else {
cout << move << " " << count << "\n";
}
}
return 0;
}
Code(Segment Tree)
#include <bits/stdc++.h>
using namespace std;
#define int long long
int modN = 1e9 + 7;
int mod(int n) {
return (n + modN) % modN;
}
struct SegmentTree {
struct Node {
int val = 1e18;
int cnt = 1;
};
vector<Node> st;
int n;
SegmentTree(int n): n(n) {
st.resize(4 * n + 1, Node());
}
SegmentTree(vector<int> a): n(a.size()) {
st.resize(4 * n + 1, Node());
build(a, 1, 0, n - 1);
}
void merge(Node& a, Node& b, Node& c) {
a.val = min(b.val, c.val);
if (b.val == c.val)
a.cnt = mod(b.cnt + c.cnt);
else if (b.val < c.val)
a.cnt = b.cnt;
else if (b.val > c.val)
a.cnt = c.cnt;
}
void build(vector<int>& a, int id, int l, int r) {
if (l == r) {
st[id].val = a[l];
return;
}
int mid = (l + r) / 2;
build(a, id * 2, l, mid);
build(a, id * 2 + 1, mid + 1, r);
merge(st[id], st[id * 2], st[id * 2 + 1]);
}
void update(int id, int l, int r, int u, int val, int cnt) {
if (l == r) {
st[id].val = val; // or st[id].sum += val
st[id].cnt = cnt;
return;
}
int mid = (l + r) / 2;
if (u <= mid) update(id * 2, l, mid, u, val, cnt);
else update(id * 2 + 1, mid + 1, r, u, val, cnt);
merge(st[id], st[id * 2], st[id * 2 + 1]);
}
void update(int idx, int val, int cnt) { //wrapper
update(1, 0, n - 1, idx, val, cnt);
}
Node query(int id, int l, int r, int u, int v) { //give 0, n - 1 as l and r and 1 as id
if (v < l || r < u) return Node();
if (u <= l && r <= v) {
return st[id];
}
int mid = (l + r) / 2;
auto a = query(id * 2, l, mid, u, v);
auto b = query(id * 2 + 1, mid + 1, r, u, v);
Node res;
merge(res, a, b);
return res;
}
Node query(int l, int r) { //wrapper
return query(1, 0, n - 1, l, r);
}
};
void solve() {
int n, m;
cin >> n >> m;
vector<int> a(n), b(m);
for (int &A : a) cin >> A;
for (int &B : b) cin >> B;
if (*max_element(a.begin(), a.end()) > b[0]) {
cout << -1 << '\n';
return;
}
vector<vector<int>> nxt(m, vector<int>(n));
for (int i = 0; i < m; i++) {
int curr = 0, r = -1;
for (int j = 0; j < n; j++) {
while (r + 1 < n && curr + a[r + 1] <= b[i])
curr += a[r + 1], r += 1;
nxt[i][j] = r + 1;
if (j <= r) curr -= a[j];
r = max(r, j);
}
}
vector<SegmentTree> dp(m, SegmentTree(vector<int>(n + 1, 1e18)));
for (int i = 0; i < m; i++)
dp[i].update(n, 0, 1);
for (int i = n - 1; i >= 0; i--) {
for (int j = m - 1; j >= 0; j--) {
auto q1 = dp[j].query(i + 1, nxt[j][i]);
int v1 = q1.val + m - (j + 1), ps1 = q1.cnt;
if (i + 1 <= nxt[j][i]) dp[j].update(i, v1, ps1);
if (j != m - 1) {
auto q2 = dp[j + 1].query(i, i);
int v2 = q2.val, ps2 = q2.cnt;
auto q3 = dp[j].query(i, i);
if (v2 < q3.val)
dp[j].update(i, v2, ps2);
else if (v2 == q3.val)
dp[j].update(i, v2, mod(ps2 + q3.cnt));
}
}
}
cout << dp[0].query(0, 0).val << ' ' << dp[0].query(0, 0).cnt << '\n';
}
signed main() {
cin.tie(0) -> sync_with_stdio(false);
int t;
cin >> t;
while (t--)
solve();
return 0;
}
2027E1 - Битовая игра (простая версия)
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
int nimber(int x, int a) {
int aprime = 0;
bool goodbit = false;
for (int bit=30; bit>=0; bit--) {
if (x & (1 << bit)) {
aprime *= 2;
if (goodbit || (a & (1 << bit))) {
aprime += 1;
}
} else if (a & (1 << bit)) {
goodbit = true;
}
}
// g(2^k - 2) = 0, for all k >= 1.
for (int k=1; k<=30; k++) {
if (aprime == (1 << k) - 2) {
return 0;
}
}
// g(2^k - 1) = k, for all k >= 1.
for (int k=1; k<=30; k++) {
if (aprime == (1 << k) - 1) {
return k;
}
}
// g(2^k) = k + (-1)^k, for all k >= 0.
for (int k=1; k<=30; k++) {
if (aprime == (1 << k)) {
if (k % 2) return k - 1;
else return k + 1;
}
}
// g(2^k+1) = g(2^k+2) = ... = g(2^{k+1} - 3) = k + 1, for all k >= 2.
for (int k=2; k<=30; k++) {
if ((1 << k) < aprime && aprime <= (2 << k) - 3) {
return k + 1;
}
}
// should never get to this point
assert(false);
return -1;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> A(n);
for (int i=0; i<n; i++) cin >> A[i];
vector<int> X(n);
for (int i=0; i<n; i++) cin >> X[i];
int curr = 0;
for (int i=0; i<n; i++) curr ^= nimber(X[i], A[i]);
cout << (curr ? "Alice" : "Bob") << "\n";
}
return 0;
}
2027E2 - Битовая игра (сложная версия)
Tutorial
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Code
#include <bits/stdc++.h>
using namespace std;
int dp[32][32][6][2][2];
const int mod = 1000000007;
int main() {
cin.tie(0)->sync_with_stdio(0);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> A(n);
for (int i=0; i<n; i++) cin >> A[i];
vector<int> B(n);
for (int i=0; i<n; i++) cin >> B[i];
vector<int> curr(32);
curr[0] = 1; // identity
for (int i=0; i<n; i++) {
memset(dp, 0, sizeof dp);
dp[0][0][0][0][0] = 1;
for (int j=0; j<=29; j++) {
int p = 29 - j; // place we are going to add a bit in
for (int k=0; k<=29; k++) { // position of most significant one in a'
for (int type=0; type<6; type++) { // 0, 1, 11111, 11110, 10000, else
for (int good=0; good<2; good++) { // good=1 iff good bit has occured
for (int low=0; low<2; low++) { // low=1 iff prefix below b
for (int bit=0; bit<2; bit++) { // bit in x
if (dp[j][k][type][good][low] == 0)
continue; // no point in transition since count is 0
if (!low && (B[i] & (1 << p)) == 0 && bit == 1)
continue; // x can't go higher than B[i]
if (bit == 0) {
int good2 = good || (A[i] & (1 << p)) != 0; // check good bit
int low2 = low || (B[i] & (1 << p)) != 0; // check if low
// nothing added to a' so nothing else changes
(dp[j+1][k][type][good2][low2] += dp[j][k][type][good][low]) %= mod;
} else {
int bita = good || (A[i] & (1 << p)) != 0; // bit in a'
int k2 = type == 0 ? 0 : k + 1; // increase if MSOne exists
int type2 = bita ?
(
type == 0 ? 1 : // add first one
(type == 1 || type == 2) ? 2 : // 11111
5 // can't add 1 after a 0
) : (
(type == 0) ? 0 : // 0
(type == 1 || type == 4) ? 4 : // 10000
(type == 2) ? 3 : // 11110
5 // can't have a zero in any other case
);
(dp[j+1][k2][type2][good][low] += dp[j][k][type][good][low]) %= mod;
}
}
}
}
}
}
}
vector<int> count(32); // number of x-values for each nimber
for (int k=0; k<=29; k++) { // position of MSOne
for (int good=0; good<2; good++) { // doesn't matter
for (int low=0; low<2; low++) { // doesn't matter
(count[0] += dp[30][k][0][good][low]) %= mod; // 0
(count[1] += dp[30][k][1][good][low]) %= mod; // 1
(count[k+1] += dp[30][k][2][good][low]) %= mod; // 11111
(count[0] += dp[30][k][3][good][low]) %= mod; // 11110
(count[k+(k%2?-1:1)] += dp[30][k][4][good][low]) %= mod; // 10000
(count[k+1] += dp[30][k][5][good][low]) %= mod; // else
}
}
}
count[0] -= 1; // remove when x=0
vector<int> next(32); // knapsack after adding this pile
for (int j=0; j<32; j++)
for (int k=0; k<32; k++)
(next[j ^ k] += 1LL * curr[j] * count[k] % mod) %= mod;
swap(curr, next);
}
cout << curr[0] << "\n";
}
return 0;
}