結果

問題 No.2568 列辞書順列列
ユーザー Be11T_
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
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 2
other AC * 25
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#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;
}
}
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