結果
問題 | No.2738 CPC To F |
ユーザー |
|
提出日時 | 2024-04-20 17:43:37 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 17 ms / 2,000 ms |
コード長 | 3,453 bytes |
コンパイル時間 | 1,245 ms |
コンパイル使用メモリ | 117,472 KB |
最終ジャッジ日時 | 2025-02-21 07:05:42 |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 22 |
ソースコード
#include <iostream>#include <algorithm>#include <math.h>#include <vector>#include <map>#include <stack>#include <queue>#include <deque>#include <set>#include <string>#include <unordered_map>#include <unordered_set>#include <numeric>#include <iomanip>constexpr int MAX_INT = 2147483647;constexpr int MIN_INT = -2147483648;#define rep(n, a, b) for (int (n) = (a); (n) < (b); (n)++)#define rrep(n, a, b) for (int (n) = (a); (n) >= (b); (n)--)#define llrep(n, a, b) for (long long (n) = (a); (n) < (b); (n)++)#define llrrep(n, a, b) for (long long (n) = (a); (n) >= (b); (n)--)#define itrAll(x) (x).begin(), (x).end()#define inputvec(v) for (auto& x : v) {std::cin >> x;}using namespace std;using uint = unsigned int;using ushort = unsigned short;using byte = unsigned char;using ll = long long;using ull = unsigned long long;using ldouble = long double;using ipair = pair<int, int>;using llpair = pair<long long, long long>;int main() {int N;string S;cin >> N >> S;vector<ipair> cpc; // { 開始インデックス, 繰り返し回数 }rep(i, 0, N - 2) {if (S[i] == 'C') {int r_cnt = 0;for (int j = i + 1; j < N - 1; j += 2) {if (S[j] == 'P' && S[j + 1] == 'C') {r_cnt++;}else {break;}}if (r_cnt) {cpc.push_back({ i, r_cnt });i += (r_cnt << 1);}}}if (!cpc.size()) {cout << 0 << endl;return 0;}/// <summary>/// [i] := {/// [first]: i+1番目のCPC...PCまでの部分文字列のうち, 少なくともi番目の「最後の」CPCを変換したときのCPCTFの個数の最大値./// [second]: 〃のうち, i番目のCPCを変換していないときの〃./// }/// </summary>vector<ipair> cnt(cpc.size());cnt[0] = { 0,0 };for (int i = 1; i < cpc.size(); i++) {auto& prev = cpc[i - 1];auto& current = cpc[i];ipair prevIdx = { prev.first, prev.first + (prev.second << 1) }, currentIdx = { current.first, current.first + (current.second << 1) };int margin = currentIdx.first - prevIdx.second - 1;if (margin == 1 && S[currentIdx.first - 1] == 'T') { // [..CPC]T[CPC..]if (current.second <= 2) {cnt[i] = { cnt[i - 1].second + 1, max(cnt[i - 1].first, cnt[i - 1].second) };}else {int v = max(cnt[i - 1].first, cnt[i - 1].second + 1);cnt[i] = { v,v };}}else if (margin >= 2 && S[prevIdx.second + 1] == 'T' && S[prevIdx.second + 2] == 'F') { // [..CPC]TF..[CPC..]int v = max(cnt[i - 1].first, cnt[i - 1].second + 1);cnt[i] = { v,v };}else { // [..CPC][CPC..] or [..CPC]..[CPC..]int v = max(cnt[i - 1].first, cnt[i - 1].second);cnt[i] = { v,v };}}auto& lastCpc = *(cpc.end() - 1);ipair lastIdx = { lastCpc.first, lastCpc.first + (lastCpc.second << 1) };auto& c = *(cnt.end() - 1);if (lastIdx.second + 2 < N && S[lastIdx.second + 1] == 'T' && S[lastIdx.second + 2] == 'F') {cout << max(c.first, c.second + 1) << endl;}else {cout << max(c.first, c.second) << endl;}return 0;}