結果
問題 | No.2941 Sigma Music Game Score Problem |
ユーザー | 👑 binap |
提出日時 | 2024-10-18 22:53:01 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 740 ms / 2,500 ms |
コード長 | 2,596 bytes |
コンパイル時間 | 4,623 ms |
コンパイル使用メモリ | 264,188 KB |
実行使用メモリ | 11,036 KB |
最終ジャッジ日時 | 2024-10-18 22:56:40 |
合計ジャッジ時間 | 12,745 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge7 |
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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 | 1 ms
6,944 KB |
testcase_04 | AC | 2 ms
6,940 KB |
testcase_05 | AC | 2 ms
6,940 KB |
testcase_06 | AC | 2 ms
6,940 KB |
testcase_07 | AC | 2 ms
6,940 KB |
testcase_08 | AC | 2 ms
6,944 KB |
testcase_09 | AC | 5 ms
6,940 KB |
testcase_10 | AC | 2 ms
6,940 KB |
testcase_11 | AC | 4 ms
6,940 KB |
testcase_12 | AC | 3 ms
6,940 KB |
testcase_13 | AC | 6 ms
6,940 KB |
testcase_14 | AC | 7 ms
6,940 KB |
testcase_15 | AC | 3 ms
6,940 KB |
testcase_16 | AC | 521 ms
8,852 KB |
testcase_17 | AC | 603 ms
9,656 KB |
testcase_18 | AC | 657 ms
10,816 KB |
testcase_19 | AC | 531 ms
9,012 KB |
testcase_20 | AC | 461 ms
7,932 KB |
testcase_21 | AC | 187 ms
6,944 KB |
testcase_22 | AC | 205 ms
6,940 KB |
testcase_23 | AC | 190 ms
6,940 KB |
testcase_24 | AC | 666 ms
10,824 KB |
testcase_25 | AC | 78 ms
6,940 KB |
testcase_26 | AC | 2 ms
6,940 KB |
testcase_27 | AC | 2 ms
6,944 KB |
testcase_28 | AC | 2 ms
6,940 KB |
testcase_29 | AC | 183 ms
6,940 KB |
testcase_30 | AC | 281 ms
8,116 KB |
testcase_31 | AC | 740 ms
11,032 KB |
testcase_32 | AC | 717 ms
11,036 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(){ long long m; int n; cin >> m >> n; vector<long long> a(n + 2); rep(i, n) cin >> a[i + 1]; a[n + 1] = m + 1; mint ans = 0; auto f = [&](mint x){ return (x - 1) * x * (2 * x - 1) / 6; }; rep(i, n + 1){ ans += f(a[i + 1] - a[i]); } cout << ans << "\n"; return 0; }