結果
問題 | No.2568 列辞書順列列 |
ユーザー |
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提出日時 | 2023-12-02 15:05:31 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 291 ms / 3,000 ms |
コード長 | 2,565 bytes |
コンパイル時間 | 4,432 ms |
コンパイル使用メモリ | 260,260 KB |
最終ジャッジ日時 | 2025-02-18 04:14:27 |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 25 |
ソースコード
#include <bits/stdc++.h> #include <atcoder/all> using namespace atcoder; #pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #define rep(i, n) for(ll i = 0; i < n; i++) #define rep2(i, m, n) for(ll i = m; i <= n; i++) #define rrep(i, m, n) for(ll i = m; i >= n; i--) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define MAX(x) *max_element(all(x)) #define MIN(x) *min_element(all(x)) #define SZ(a) ((ll)(a).size()) #define bit(n, k) ((n >> k) & 1) using namespace std; using ll = long long; using P = pair<ll, ll>; const int dx[4] = { 1, 0, -1, 0}; const int dy[4] = { 0, -1, 0, 1}; const int inf = 1e9 + 7; const ll infll = 1e18; //const double pi = acos(-1); using Graph = vector<vector<ll>>; using REV_PQ = priority_queue<ll, vector<ll>, greater<ll>>; using PQ = priority_queue<ll>; const int SIZE = 400010; ll powll(ll n, ll x) { // return n ^ x; ll ret = 1; rep(i, x) ret *= n; return ret; } void OutputYesNo(bool val) { if (val) cout << "Yes" << endl; else cout << "No" << endl; } void OutputTakahashiAoki(bool val) { if (val) cout << "Takahashi" << endl; else cout << "Aoki" << endl; } void OutPutInteger(ll x) { cout << x << endl; } void OutPutString(string x) { cout << x << endl; } int pop_count(ll n){ int ret = 0; while(n > 0){ if(n % 2 == 1) ret++; n /= 2; } return ret; } using PAIR_REV_PQ = priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>>>; typedef atcoder::modint998244353 mint; //typedef atcoder::modint1000000007 mint; //typedef modint mint; //typedef vector<mint> ModintVec; mint beki[200010]; mint x[200010]; struct S{ mint sum; ll len; }; S op(S a,S b){ return {a.sum*beki[b.len]+b.sum,a.len+b.len}; } S e(){ return {0,0}; } using F = ll; S mapping(F f,S s){ if(f==0) return s; return {f*x[s.len],s.len}; } F composition(F f,F g){ if(f==0) return g; return f; } F id(){ return 0; } int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); ll n, m, q; cin >> n >> m >> q; vector<ll> a(n); rep(i, n) cin >> a[i]; rep(i, n) a[i]--; vector<S> v(n); rep(i, n) v[i] = {a[i], 1}; beki[0] = 1; rep(i, 200000){ beki[i + 1] = beki[i] * m; x[i + 1] = x[i] + beki[i]; } atcoder::lazy_segtree<S,op,e,F,mapping,composition,id> seg(v); while(q--){ ll l, r; cin >> l >> r; l--; mint res = seg.prod(l, r).sum.val(); res++; cout << res.val() << endl; } }