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 leaf_1415
提出日時 2021-06-11 22:53:51
言語 C++11
(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
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
other AC * 42
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 <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); //
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;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0