結果
問題 | No.2281 K → K-1 01 Flip |
ユーザー |
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提出日時 | 2023-04-20 10:00:59 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 2,685 bytes |
コンパイル時間 | 2,057 ms |
コンパイル使用メモリ | 177,432 KB |
実行使用メモリ | 14,464 KB |
最終ジャッジ日時 | 2024-10-15 10:39:54 |
合計ジャッジ時間 | 24,037 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 6 WA * 50 |
ソースコード
#include <bits/stdc++.h>using namespace std;template <typename T>struct SegTree {const T INF = numeric_limits<T>::max();int n;vector<T> dat;SegTree(int n_) : dat(n_ * 4, INF) {int x = 1;while (n_ > x) {x *= 2;}n = x;}void init(T x) {for (int i = 0; i < 2 * n - 1; i++) dat[i] = x;}void update(int i, T x) {i += n - 1;dat[i] = x;while (i > 0) {i = (i - 1) / 2;dat[i] = max(dat[i * 2 + 1], dat[i * 2 + 2]);}}T query(int a, int b) {return query_sub(a, b, 0, 0, n);}T query_sub(int a, int b, int k, int l, int r) {if (r <= a || b <= l) {return INF;}else if (a <= l && r <= b) {return dat[k];}else {T vl = query_sub(a, b, k * 2 + 1, l, (l + r) / 2);T vr = query_sub(a, b, k * 2 + 2, (l + r) / 2, r);return max(vl, vr);}}};int main() {long long N, Q; cin >> N >> Q;string S; cin >> S;vector<pair<long long, long long>> rle;vector<long long> a(N);vector<long long> left;for (int i = 0; i < N; i++) {if (i == 0) {if (S[i] == '0') rle.push_back(make_pair(0, 1));if (S[i] == '1') rle.push_back(make_pair(1, 1));}else {if (rle[rle.size() - 1].first == 0 && S[i] == '0') rle[rle.size() - 1].second++;else if (rle[rle.size() - 1].first == 1 && S[i] == '0') rle.push_back(make_pair(0, 1));else if (rle[rle.size() - 1].first == 0 && S[i] == '1') rle.push_back(make_pair(1, 1));else if (rle[rle.size() - 1].first == 1 && S[i] == '1') rle[rle.size() - 1].second++;}a[i] = rle.size() - 1;if (rle[rle.size() - 1].second == 1) left.push_back(i);}SegTree<long long> seg(rle.size());for (int i = 0; i < rle.size(); i++) {seg.update(i, rle[i].second);}vector<long long> rui_l(N, 0), rui_r(N, 0);for (int i = 0; i < N; i++) {if (i != 0) rui_l[i] = rui_l[i - 1];if (S[i] == '0') rui_l[i] += 1;else rui_l[i] -= 1;if (i != 0) rui_r[N - i - 1] = rui_r[N - i];if (S[N - i - 1] == '0') rui_r[N - i - 1] += 1;else rui_r[N - i - 1] -= 1;}for (int i = 0; i < Q; i++) {long long L, R, K; cin >> L >> R >> K;long long max_ = 0;if (a[R - 1] > a[L - 1] + 1) max_ = seg.query(a[L - 1] + 1, a[R - 1]);if (a[L - 1] + 1 < rle.size()) max_ = max(max_, left[a[L - 1] + 1] - (L - 1));max_ = max(max_, (R - 1) - left[a[R - 1]] + 1);if (max_ < K) cout << R - L + 1 << endl;else {long long sum = rui_l[N - 1];if (L != 1) sum -= rui_l[L - 2];if (R != N) sum -= rui_r[R];long long amari = (sum + 1000000 * (2 * K - 1)) % (2 * K - 1);if (amari == 0) cout << 2 * (K - 1) << endl;else if (amari < K) cout << 2 * K - 2 - amari << endl;else cout << amari - 1 << endl;}}}