結果
問題 | No.1548 [Cherry 2nd Tune B] 貴方と私とサイクルとモーメント |
ユーザー |
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提出日時 | 2021-06-11 22:53:51 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 252 ms / 4,500 ms |
コード長 | 8,542 bytes |
コンパイル時間 | 3,366 ms |
コンパイル使用メモリ | 99,404 KB |
実行使用メモリ | 37,604 KB |
最終ジャッジ日時 | 2024-12-15 01:42:41 |
合計ジャッジ時間 | 11,643 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 42 |
ソースコード
#include <iostream> #include <cstdio> #include <cmath> #include <ctime> #include <cstdlib> #include <cassert> #include <vector> #include <list> #include <stack> #include <queue> #include <deque> #include <map> #include <set> #include <bitset> #include <string> #include <algorithm> #include <utility> #include <complex> #define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++) #define per(x, s, t) for(llint (x) = (s); (x) >= (t); (x)--) #define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++) #define chmin(x, y) (x) = min((x), (y)) #define chmax(x, y) (x) = max((x), (y)) #define sz(x) ((ll)(x).size()) #define ceil(x, y) (((x)+(y)-1) / (y)) #define all(x) (x).begin(),(x).end() #define outl(...) dump_func(__VA_ARGS__) #define inf 1e18 using namespace std; typedef long long llint; typedef long long ll; typedef pair<ll, ll> P; struct edge{ ll to, cost; edge(){} edge(ll a, ll b){ to = a, cost = b;} }; const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1}; //const ll mod = 1000000007; const ll mod = 998244353; struct mint{ ll x = 0; mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;} mint(const mint &ope) {x = ope.x;} mint operator-(){return mint(-x);} mint operator+(const mint &ope){return mint(x) += ope;} mint operator-(const mint &ope){return mint(x) -= ope;} mint operator*(const mint &ope){return mint(x) *= ope;} mint operator/(const mint &ope){return mint(x) /= ope;} mint& operator+=(const mint &ope){ x += ope.x; if(x >= mod) x -= mod; return *this; } mint& operator-=(const mint &ope){ x += mod - ope.x; if(x >= mod) x -= mod; return *this; } mint& operator*=(const mint &ope){ x *= ope.x, x %= mod; return *this; } mint& operator/=(const mint &ope){ ll n = mod-2; mint mul = ope; while(n){ if(n & 1) *this *= mul; mul *= mul; n >>= 1; } return *this; } mint inverse(){return mint(1) / *this;} bool operator ==(const mint &ope){return x == ope.x;} bool operator !=(const mint &ope){return x != ope.x;} }; mint modpow(mint a, ll n){ if(n == 0) return mint(1); if(n % 2) return a * modpow(a, n-1); else return modpow(a*a, n/2); } istream& operator >>(istream &is, mint &ope){ ll t; is >> t, ope.x = t; return is; } ostream& operator <<(ostream &os, mint &ope){return os << ope.x;} ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;} bool exceed(ll x, ll y, ll m){return x >= m / y + 1;} void mark(){ cout << "*" << endl; } void yes(){ cout << "YES" << endl; } void no(){ cout << "NO" << endl; } ll sgn(ll x){ if(x > 0) return 1; if(x < 0) return -1; return 0;} ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);} ll lcm(ll a, ll b){return a/gcd(a, b)*b;} ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;} ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;} string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;} ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;} template<typename T> void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());} template<typename T> ostream& operator << (ostream& os, vector<T>& vec) { for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " "); return os; } template<typename T> ostream& operator << (ostream& os, deque<T>& deq) { for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " "); return os; } template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) { os << "(" << pair_var.first << ", " << pair_var.second << ")"; return os; } template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U>& pair_var) { os << "(" << pair_var.first << ", " << pair_var.second << ")"; return os; } template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) { for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) { os << "(" << itr->first << ", " << itr->second << ")"; itr++; if(itr != map_var.end()) os << ","; itr--; } return os; } template<typename T> ostream& operator << (ostream& os, set<T>& set_var) { for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) { os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--; } return os; } template<typename T> ostream& operator << (ostream& os, multiset<T>& set_var) { for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) { os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--; } return os; } template<typename T> void outa(T a[], ll s, ll t){ for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";} cout << endl; } void dump_func() {cout << endl;} template <class Head, class... Tail> void dump_func(Head &&head, Tail &&... tail) { cout << head; if(sizeof...(Tail) > 0) cout << " "; dump_func(std::move(tail)...); } struct A{ mint a[4]; A(){a[0] = a[1] = a[2] = a[3] = mint(0);} A(mint x){ a[0] = x; rep(i, 1, 3) a[i] = a[i-1] * x; } A operator+(const A &ope){ A ret; rep(i, 0, 3) ret.a[i] = a[i] + ope.a[i]; return ret; } A operator*(const ll x){ A ret; rep(i, 0, 3) ret.a[i] = a[i]*x; return ret; } }; struct LazySegTree{ typedef A SEG; typedef A DELAY; int size; vector<SEG> seg; vector<DELAY> delay; LazySegTree(){} LazySegTree(int size){ this->size = size; seg.resize(1<<(size+1)); delay.resize(1<<(size+1)); } SEG Ident(){ //identity element return A(); } SEG ope(SEG a, SEG b){ //operator return a+b; } void init() { for(int i = 0; i < (1<<(size+1)); i++){ seg[i] = Ident(); delay[i] = A(); // } } void eval(int l, int r, int k) // { if(delay[k].a[0] != mint(0)){ seg[k] = delay[k] * (r-l+1); //総和クエリのときは区間幅を乗じる必要がある if(l < r){ delay[k*2] = delay[k]; delay[k*2+1] = delay[k]; } delay[k] = A(); } } void update(int i, SEG val) { int l = 0, r = (1<<size)-1, k = 1; eval(l, r, k); for(int j = size-1; j >= 0; j--){ k <<= 1; if(i & (1<<j)){ k++; l = (l+r)/2+1; } else r = (l+r)/2; eval(l, r, k); } seg[i+(1<<size)] = val; l = i, r = i, k = i+(1<<size); for(int j = 0; j < size; j++){ k /= 2, l &= ~(1<<j), r |= 1<<j; eval(l, (l+r)/2, k*2), eval((l+r)/2+1, r, k*2+1); seg[k] = ope(seg[k*2], seg[k*2+1]); } } void add(int a, int b, int k, int l, int r, DELAY val) { eval(l, r, k); if(b < l || r < a) return; if(a <= l && r <= b){ delay[k] = val; // eval(l, r, k); return; } add(a, b, k*2, l, (l+r)/2, val); add(a, b, k*2+1, (l+r)/2+1, r, val); seg[k] = ope(seg[k*2], seg[k*2+1]); } void add(int a, int b, DELAY val){ if(a > b) return; add(a, b, 1, 0, (1<<size)-1, val); } SEG query(int a, int b, int k, int l, int r) { eval(l, r, k); if(b < l || r < a) return Ident(); if(a <= l && r <= b) return seg[k]; SEG lval = query(a, b, k*2, l, (l+r)/2); SEG rval = query(a, b, k*2+1, (l+r)/2+1, r); return ope(lval, rval); } SEG query(int a, int b) { if(a > b) return Ident(); return query(a, b, 1, 0, (1<<size)-1); } }; ll n, Q; ll a[200005]; LazySegTree seg(18); vector<P> get(ll u, ll v, ll w) { if(u > v) swap(u, v); vector<P> ret; if(u < w && w < v) ret.push_back(P(u, v)); else{ ret.push_back(P(1, u)); ret.push_back(P(v, n)); } return ret; } int main(void) { ios::sync_with_stdio(0); cin.tie(0); cin >> n; rep(i, 1, n) cin >> a[i]; seg.init(); rep(i, 1, n) seg.update(i, A(mint(a[i]))); cin >> Q; ll t, u, v, w, b; rep(q, 1, Q){ cin >> t >> u >> v >> w; vector<P> vec = get(u, v, w); if(t == 0){ cin >> b; for(auto p : vec){ seg.add(p.first, p.second, A(mint(b))); } continue; } ll len = 0; A res; for(auto p : vec){ len += p.second - p.first + 1; res = res + seg.query(p.first, p.second); } mint m = res.a[0] / mint(len), ans = mint(0); if(t == 1){ ans = res.a[0] - mint(len) * m; } if(t == 2){ ans = res.a[1] - mint(2) * res.a[0] * m + mint(len) * m * m; } if(t == 3){ ans = res.a[2] - mint(3) * res.a[1] * m + mint(3) * res.a[0] * m * m - mint(len) * m * m * m; } if(t == 4){ ans = res.a[3] - mint(4) * res.a[2] * m + mint(6) * res.a[1] * m * m - mint(4) * res.a[0] * m * m * m + mint(len) * m * m * m * m; } ans /= mint(len); outl(ans); } return 0; }