Increase/Decrease/Copy
我们可以先将 \(a_i\) 变为 \(b_i\),统计在变化的过程中与 \(b_{i + 1}\)的最少差值即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 2e5 + 5;int t, n, a[N], b[N];void Solve() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i];}for (int i = 1; i <= n + 1; i++) {cin >> b[i];}int mini = 1e9, ans = 0;for (int i = 1; i <= n; i++) {if ((a[i] <= b[n + 1] && b[n + 1] <= b[i]) || (b[i] <= b[n + 1] && b[n + 1] <= a[i])) {mini = 0;}mini = min(mini, abs(a[i] - b[n + 1]));mini = min(mini, abs(b[i] - b[n + 1]));ans += abs(a[i] - b[i]);}cout << ans + mini + 1 << "\n";
}signed main() {ios::sync_with_stdio(0);cin.tie(0);cin >> t;while (t--) {Solve();}return 0;
}
Job Interview
大模拟,维护 \(4\) 个前缀和然后二分
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 4e5 + 5;int t, n, m, a[N], b[N], sum[3][N], cnt[3][N], arr[3][N], tot[3], cur;void Solve() {cin >> n >> m;for (int i = 1; i <= n + m + 1; i++) {cin >> a[i];}for (int i = 1; i <= n + m + 1; i++) {cin >> b[i];}for (int i = 1; i <= n + m + 1; i++) {sum[1][i] = sum[1][i - 1] + (a[i] > b[i]) * a[i];sum[2][i] = sum[2][i - 1] + (b[i] > a[i]) * b[i];cnt[1][i] = cnt[1][i - 1] + (a[i] > b[i]);cnt[2][i] = cnt[2][i - 1] + (b[i] > a[i]);arr[1][i] = arr[1][i - 1] + a[i];arr[2][i] = arr[2][i - 1] + b[i];}for (int i = 1; i <= n + m + 1; i++) {if (i == n + m + 1) {cout << cur << "\n";break;}int tmp = n + m + 1;for (auto op : {1, 2}) {int l = i, r = n + m + 1;while (l < r) {int mid = (l + r + 1) >> 1;if (cnt[op][mid] - cnt[op][i] <= n * (op == 1) + m * (op == 2) - tot[op]) {l = mid;}else r = mid - 1;}// cout << l << " ";tmp = min(tmp, l);}// cout << "\n";int now = sum[1][tmp] - sum[1][i] + sum[2][tmp] - sum[2][i];if (cnt[1][tmp] - cnt[1][i] == n - tot[1]) {cout << cur + now + arr[2][n + m + 1] - arr[2][tmp] << " ";}else {cout << cur + now + arr[1][n + m + 1] - arr[1][tmp] << " ";}if (tot[2] == m || (a[i] > b[i] && tot[1] < n)) {tot[1]++, cur += a[i];}else {tot[2]++, cur += b[i];}}tot[1] = tot[2] = cur = 0;
}signed main() {ios::sync_with_stdio(0);cin.tie(0);cin >> t;while (t--) {Solve();}return 0;
}
Invertible Bracket Sequences
用一个小根堆,考虑如果当前的和为 \(sum\) 那么之前的 \(sum1 * 2 < sum\) 的都不可以,我们开一个堆,每次统计为\(sum\)有多少个即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 4e5 + 5;int t, n, a[N], ans, cnt[N];string s;void Solve() {priority_queue<int, vector<int>, greater<int> > q;cin >> s;n = s.size();s = " " + s;for (int i = 1; i <= n + 1; i++) {cnt[i] = 0;}cnt[0] = 1;ans = 0;q.push(0);for (int i = 1; i <= n; i++) {if (s[i] == '(') {a[i] = 1;}else a[i] = -1;}for (int i = 1, sum = 0; i <= n; i++) {sum += a[i];while (!q.empty() && q.top() * 2 < sum) {cnt[q.top()]--;q.pop();}ans += cnt[sum];q.push(sum);cnt[sum]++;}cout << ans << "\n";
}signed main() {ios::sync_with_stdio(0);cin.tie(0);cin >> t;while (t--) {Solve();}return 0;
}
