Back to School '24 P1 - Kicking
按照题意模拟即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 1e5 + 5;int n, m, k;char c;vector<int> sum[2][N];vector<int> a[N];signed main() {cin >> n >> m >> k;for (int i = 1; i <= n; i++) {for (int j = 1; j <= m + 1; j++) {sum[0][i].push_back(0);sum[1][i].push_back(0);}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {cin >> c;a[i].push_back(c);sum[0][i][j] = sum[0][i][j - 1] + (c == 'A');sum[1][i][j] = sum[1][i][j - 1] + (c == 'B');}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {if (a[i][j - 1] == '.') {cout << ".";}else if (a[i][j - 1] == 'A') {if (sum[1][i][min(j + k, m)] - sum[1][i][j - 1] > 0) {cout << "N";}else cout << "Y";}else {if (sum[0][i][j] - sum[0][i][max(0ll, j - k - 1)] > 0) {cout << "N";}else cout << "Y";}}cout << "\n";}return 0;
}Back to School '24 P2 - Cheating
注意到一个重要的性质,就是 \(a_i\) 为递增序列,那么我们会发现,他改的区间一定是上一个区间后面的,模拟即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 1e6 + 5;int n, m, a[N], b[N], ans[N];signed main() {cin >> n >> m;for (int i = 1; i <= n; i++) {cin >> a[i];}for (int i = 1; i <= n; i++) {cin >> b[i];}int last = 1;for (int i = 1; i <= n; i++) {if (last + b[i] - 1 <= m) {ans[last] += a[i];ans[last + b[i]] -= a[i];last += b[i];if (last == m + 1) {last = 1;}}else {ans[last] += a[i];ans[m + 1] -= a[i];b[i] -= (m - last + 1);ans[1] += a[i];ans[min(b[i] + 1, last)] -= a[i];last = min(b[i] + 1, last);}}for (int i = 1; i <= m; i++) {ans[i] += ans[i - 1];}for (int i = 1; i <= m; i++) {cout << ans[i] << " ";}return 0;
}
Back to School '24 P3 - Tournament
我们可以发现,他合并答案是简直和线段树一摸一样,所以直接线段树维护即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = (1 << 18) + 5;struct node {int maxi, ans;void clean() {maxi = ans = 0;}
}tr[N * 4];int n, p, a[N];bool vis[N];void build(int i, int l, int r) {if (l == r) {if (l == 1) {tr[i].maxi = p, tr[i].ans = 0;}else tr[i].maxi = a[l - 1], tr[i].ans = 0;return ;}int mid = (l + r) >> 1;build(i * 2, l, mid);build(i * 2 + 1, mid + 1, r);tr[i].ans = tr[i * 2].ans + tr[i * 2 + 1].ans + (tr[i * 2].maxi - tr[i * 2 + 1].maxi) * (tr[i * 2].maxi - tr[i * 2 + 1].maxi);tr[i].maxi = max(tr[i * 2].maxi, tr[i * 2 + 1].maxi);
}void modify(int i, int l, int r, int p, int x) {if (l == r) {tr[i].maxi = x;tr[i].ans = 0;return ;}int mid = (l + r) >> 1;if (mid >= p) modify(i * 2, l, mid, p, x);else modify(i * 2 + 1, mid + 1, r, p, x);tr[i].ans = tr[i * 2].ans + tr[i * 2 + 1].ans + (tr[i * 2].maxi - tr[i * 2 + 1].maxi) * (tr[i * 2].maxi - tr[i * 2 + 1].maxi);tr[i].maxi = max(tr[i * 2].maxi, tr[i * 2 + 1].maxi);
}signed main() {cin >> n;for (int i = 1; i < n; i++) {cin >> a[i];vis[a[i]] = true;}for (int i = 1; i <= n; i++) {if (!vis[i]) {p = i;}}build(1, 1, n);for (int i = 1; i <= n; i++) {cout << tr[1].ans << " ";if (i < n) {modify(1, 1, n, i, a[i]);modify(1, 1, n, i + 1, p);}}return 0;
}
Back to School '24 P4 - Candidates
我们假如考虑每一种因素,他一定要满足在每个区间内是递增的,用 \(set\) 维护,做一个类似于拓扑排序的即可
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 5;int n, k, a[N], cnt[N];vector<int> r[N], v[N], ans;bool vis[N];priority_queue<int, vector<int>, greater<int> > q;int main() {ios::sync_with_stdio(0);cin.tie(0);cin >> n >> k;for (int i = 1; i <= n; i++) {cin >> a[i];r[i].push_back(0);}for (int i = 1; i <= n; i++) {for (int j = 1, c; j <= k; j++) {cin >> c;r[i].push_back(c);}}for (int j = 1; j <= k; j++) {for (int i = 1; i < n; i++) {if (r[a[i]][j] < r[a[i + 1]][j]) {cnt[j]++;vis[i] = true;v[i].push_back(j);}}}for (int i = 1; i <= k; i++) {if (!cnt[i]) {q.push(i);}}while (!q.empty()) {int cur = q.top();q.pop();ans.push_back(cur);vector<int> tmp;for (int i = 1; i < n; i++) {if (r[a[i]][cur] > r[a[i + 1]][cur] && vis[i]) {tmp.push_back(i);}}for (auto cur : tmp) {for (auto u : v[cur]) {cnt[u]--;if (!cnt[u]) {q.push(u);}}vis[cur] = false;v[cur].clear();}}for (int i = 1; i <= k; i++) {if (cnt[i]) {cout << "-1";return 0;}}for (auto cur : ans) {cout << cur << " ";}return 0;
}
