結果
問題 | No.2943 Sigma String of String Score Problem |
ユーザー | 👑 binap |
提出日時 | 2024-10-18 22:16:38 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 394 ms / 3,000 ms |
コード長 | 2,655 bytes |
コンパイル時間 | 5,021 ms |
コンパイル使用メモリ | 265,172 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-10-18 22:49:06 |
合計ジャッジ時間 | 9,481 ms |
ジャッジサーバーID (参考情報) |
judge8 / judge5 |
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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
6,816 KB |
testcase_02 | AC | 2 ms
6,940 KB |
testcase_03 | AC | 325 ms
6,940 KB |
testcase_04 | AC | 13 ms
6,940 KB |
testcase_05 | AC | 151 ms
6,944 KB |
testcase_06 | AC | 10 ms
6,944 KB |
testcase_07 | AC | 37 ms
6,940 KB |
testcase_08 | AC | 266 ms
6,940 KB |
testcase_09 | AC | 137 ms
6,944 KB |
testcase_10 | AC | 38 ms
6,940 KB |
testcase_11 | AC | 8 ms
6,944 KB |
testcase_12 | AC | 59 ms
6,940 KB |
testcase_13 | AC | 20 ms
6,944 KB |
testcase_14 | AC | 14 ms
6,944 KB |
testcase_15 | AC | 22 ms
6,940 KB |
testcase_16 | AC | 15 ms
6,944 KB |
testcase_17 | AC | 6 ms
6,940 KB |
testcase_18 | AC | 184 ms
6,940 KB |
testcase_19 | AC | 95 ms
6,940 KB |
testcase_20 | AC | 60 ms
6,940 KB |
testcase_21 | AC | 190 ms
6,944 KB |
testcase_22 | AC | 365 ms
6,940 KB |
testcase_23 | AC | 365 ms
6,944 KB |
testcase_24 | AC | 365 ms
6,944 KB |
testcase_25 | AC | 329 ms
6,940 KB |
testcase_26 | AC | 368 ms
6,940 KB |
testcase_27 | AC | 394 ms
6,940 KB |
testcase_28 | AC | 2 ms
6,940 KB |
testcase_29 | AC | 2 ms
6,940 KB |
testcase_30 | AC | 2 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; 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 <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;} template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;} 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);} using mint = modint998244353; int main(){ string s, t; cin >> s >> t; mint ans = 0; int ns = s.size(); int nt = t.size(); vector<mint> dp(nt + 1); dp[0] = 1; rep(i, ns){ vector<mint> dp_old(nt + 1); swap(dp, dp_old); rep(j, nt + 1) dp[j] += dp_old[j] * 2; rep(j, nt){ if(s[i] == t[j]) dp[j + 1] += dp_old[j]; } } cout << dp[nt] << "\n"; return 0; }