結果
問題 | No.2894 Monotonic Intervals |
ユーザー | 👑 binap |
提出日時 | 2024-09-20 22:45:34 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 22 ms / 2,000 ms |
コード長 | 3,215 bytes |
コンパイル時間 | 4,605 ms |
コンパイル使用メモリ | 267,436 KB |
実行使用メモリ | 8,892 KB |
最終ジャッジ日時 | 2024-09-20 22:45:45 |
合計ジャッジ時間 | 5,672 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,812 KB |
testcase_01 | AC | 2 ms
6,944 KB |
testcase_02 | AC | 2 ms
6,944 KB |
testcase_03 | AC | 2 ms
6,940 KB |
testcase_04 | AC | 17 ms
6,940 KB |
testcase_05 | AC | 2 ms
6,944 KB |
testcase_06 | AC | 2 ms
6,940 KB |
testcase_07 | AC | 10 ms
6,944 KB |
testcase_08 | AC | 15 ms
6,940 KB |
testcase_09 | AC | 19 ms
8,572 KB |
testcase_10 | AC | 16 ms
6,976 KB |
testcase_11 | AC | 22 ms
8,888 KB |
testcase_12 | AC | 21 ms
8,892 KB |
testcase_13 | AC | 17 ms
6,944 KB |
testcase_14 | AC | 17 ms
6,940 KB |
testcase_15 | AC | 17 ms
6,940 KB |
testcase_16 | AC | 17 ms
6,948 KB |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> #define rep(i,n) for(int i=0;i<n;i++) using namespace std; using namespace atcoder; typedef long long ll; typedef vector<int> vi; typedef vector<long long> vl; typedef vector<vector<int>> vvi; typedef vector<vector<long long>> vvl; typedef long double ld; typedef pair<int, int> P; ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;} template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;} template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;} template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;} template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename T> void chmin(T& a, T b){a = min(a, b);} template<typename T> void chmax(T& a, T b){a = max(a, b);} // thanks for snuke-san's blog // https://snuke.hatenablog.com/entry/2014/12/02/235837 vector<int> manacher(vector<int>& s){ int n = s.size(); vector<int> res(n); int i = 0, j = 0; while(i < n){ while(i - j >= 0 && i + j < n && s[i-j] == s[i+j]) ++j; res[i] = j; int k = 1; while(i - k >= 0 && k + res[i-k] < j) res[i+k] = res[i-k], ++k; i += k; j -= k; } return res; } vector<int> manacher(string& s){ vector<int> s_converted; for(char c : s) s_converted.push_back(c); return manacher(s_converted); } // https://atcoder.jp/contests/abc259/editorial/4283 vector<pair<char, int>> RLE(vector<int>& s){ vector<pair<char, int>> res; int cnt = 1; for(int i = 1; i < (int)s.size(); i++){ if(s[i] != s[i-1]){ res.push_back({s[i-1], cnt}); cnt = 0; } cnt++; } res.push_back({s.back(), cnt}); return res; } vector<pair<char, int>> RLE(string& s){ vector<int> s_converted; for(char c : s) s_converted.push_back(c); return RLE(s_converted); } int main(){ int n; cin >> n; string s; cin >> s; auto rle = RLE(s); cout << rle.size() << "\n"; return 0; }