結果
| 問題 |
No.3070 Collecting Coins Speedrun 2
|
| コンテスト | |
| ユーザー |
👑 |
| 提出日時 | 2025-03-21 21:24:14 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 45 ms / 2,000 ms |
| コード長 | 2,475 bytes |
| コンパイル時間 | 4,204 ms |
| コンパイル使用メモリ | 250,340 KB |
| 実行使用メモリ | 5,888 KB |
| 最終ジャッジ日時 | 2025-03-21 21:24:22 |
| 合計ジャッジ時間 | 6,920 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 32 |
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 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(){
int n;
cin >> n;
vector<int> a(n);
cin >> a;
if(n == 1){
cout << "1\n";
return 0;
}
mint ans = 1;
if(a[0] < 0 and a[n - 1] > 0){
ans *= 2;
for(int i = 1; i < n - 1; i++){
if(a[i] == 0) ans *= 3;
else ans *= 2;
}
}else{
ans = mint(2).pow(n - 1);
}
cout << ans << "\n";
return 0;
}