QMAX2 - Giá trị lớn nhất ver2
Tác giả: RR
Ngôn ngữ: C++
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; ++i)
#define FORD(i,a,b) for(int i=(a),_b=(b); i>=_b; --i)
#define REP(i,a) for(int i=0,_a=(a); i < _a; ++i)
#define DEBUG(X) { cout << #X << " = " << (X) << endl; }
#define PR(A,n) { cout << #A << " = "; FOR(_,1,n) cout << A[_] << ' '; cout << endl; }
#define PR0(A,n) { cout << #A << " = "; REP(_,n) cout << A[_] << ' '; cout << endl; }
#define sqr(x) ((x) * (x))
#define ll long long
#define __builtin_popcount __builtin_popcountll
#define SZ(x) ((int) (x).size())
using namespace std;
double safe_sqrt(double x) {
return sqrt(max((double)0.0,x));
}
int GI(ll& x) {
return scanf("%lld", &x);
}
const int MAXN = 50111;
struct Node {
int lazy; // giá trị T trong phân tích trên
int val; // giá trị lớn nhất
} nodes[MAXN * 4];
void down(int id) {
int t = nodes[id].lazy;
nodes[id*2].lazy += t;
nodes[id*2].val += t;
nodes[id*2+1].lazy += t;
nodes[id*2+1].val += t;
nodes[id].lazy = 0;
}
void update(int id, int l, int r, int u, int v, int val) {
if (v < l || r < u) {
return ;
}
if (u <= l && r <= v) {
nodes[id].val += val;
nodes[id].lazy += val;
return ;
}
int mid = (l + r) / 2;
down(id);
update(id*2, l, mid, u, v, val);
update(id*2+1, mid+1, r, u, v, val);
nodes[id].val = max(nodes[id*2].val, nodes[id*2+1].val);
}
int get(int id, int l, int r, int u, int v) {
if (v < l || r < u) {
return -2000111000;
}
if (u <= l && r <= v) {
return nodes[id].val;
}
int mid = (l + r) / 2;
down(id);
return max(get(id*2, l, mid, u, v),
get(id*2+1, mid+1, r, u, v));
}
int main() {
ios :: sync_with_stdio(0); cin.tie(0);
int n, q; cin >> n >> q;
while (q--) {
int typ; cin >> typ;
if (typ == 0) {
int l, r, val; cin >> l >> r >> val;
update(1, 1, n, l, r, val);
}
else {
int l, r; cin >> l >> r;
printf("%d\n", get(1, 1, n, l, r));
}
}
}