結果

問題 No.1099 Range Square Sum
ユーザー 👑 NachiaNachia
提出日時 2020-10-03 19:05:55
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 940 ms / 2,000 ms
コード長 3,229 bytes
コンパイル時間 1,747 ms
コンパイル使用メモリ 176,560 KB
実行使用メモリ 61,932 KB
最終ジャッジ日時 2024-07-18 04:47:45
合計ジャッジ時間 12,299 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 30
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>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
#define rep(i,n) for(int i=0; i<(n); i++)
template<
class S,
S(*op)(S a, S b),
S(*e)(),
class F,
S(*mapping)(F a, S b),
F(*composition)(F a, F b),
F(*id)()
>
struct lazy_segtree {
private:
struct Node { S v; F r; };
int N;
vector<Node> V;
void spread(int i) {
V[i].v = mapping(V[i].r, V[i].v);
if (i < N) {
V[i * 2].r = composition(V[i].r, V[i * 2].r);
V[i * 2 + 1].r = composition(V[i].r, V[i * 2 + 1].r);
}
V[i].r = id();
}
public:
lazy_segtree(int n) {
N = 1; while (N < n) N *= 2;
V.assign(N * 2, { e(),id() });
}
lazy_segtree(vector<S> A) {
N = 1; while (N < A.size()) N *= 2;
V.assign(N * 2, { e(),id() });
rep(i, A.size()) V[N + i].v = A[i];
for (int i = N - 1; i >= 1; i--)
V[i].v = op(V[i * 2].v, V[i * 2 + 1].v);
}
void set(int p, S v) {
p += N;
for (int d = N; d >= 1; d /= 2) spread(p / d);
V[p].v = v;
int z = 1;
while (p != 1) {
p /= 2; z *= 2;
spread(p * 2); spread(p * 2 + 1);
V[p].v = op(V[p * 2].v, V[p * 2 + 1].v);
}
}
S get(int p) {
p += N;
for (int d = N; d >= 1; d /= 2) spread(p / d);
return V[p].v;
}
void apply(int l, int r, F v, int a = 0, int b = 0, int i = -1) {
if (i == -1) { a = 0; b = N; i = 1; }
if (r <= a || b <= l) { spread(i); return; }
else if (l <= a && b <= r) { V[i].r = composition(v, V[i].r); spread(i); return; }
spread(i);
apply(l, r, v, a, (a + b) / 2, i * 2);
apply(l, r, v, (a + b) / 2, b, i * 2 + 1);
V[i].v = op(V[i * 2].v, V[i * 2 + 1].v);
}
S prod(int l, int r, int a = 0, int b = 0, int i = -1) {
if (i == -1) { a = 0; b = N; i = 1; }
if (r <= a || b <= l) return e();
spread(i);
if (l <= a && b <= r) return V[i].v;
S q1 = prod(l, r, a, (a + b) / 2, i * 2);
S q2 = prod(l, r, (a + b) / 2, b, i * 2 + 1);
q1 = op(q1, q2);
return q1;
}
};
struct S { LL m[3] = {}; };
S op(S l, S r) { S res; rep(i, 3) res.m[i] = l.m[i] + r.m[i]; return res; }
S e() { return { {0,0,0} }; }
struct F { LL m[3][3] = {}; };
S mapping(F f, S a) { S res; rep(i, 3) rep(j, 3) res.m[i] += a.m[j] * f.m[j][i]; return res; }
F composition(F f, F g) { F res; rep(i, 3) rep(j, 3) rep(k, 3) res.m[i][j] += g.m[i][k] * f.m[k][j]; return res; }
F id() { return { {{1,0,0},{0,1,0},{0,0,1}} }; }
using RQ = lazy_segtree<S, op, e, F, mapping, composition, id>;
int main() {
int N; cin >> N;
vector<S> rqinit(N);
rep(i, N) { LL a; cin >> a; rqinit[i] = S{ {a * a,a,1} }; }
RQ G(rqinit);
int Q; cin >> Q;
rep(i, Q) {
int c; cin >> c;
if (c == 1) {
int l, r; LL x; cin >> l >> r >> x; l--;
G.apply(l, r, F{ {{1,0,0},{2 * x,1,0},{x*x,x,1}} });
}
else if (c == 2) {
int l, r; cin >> l >> r; l--;
cout << G.prod(l, r).m[0] << endl;
}
}
return 0;
}
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