#include <bits/stdc++.h>
// same as the nlog2 solution, except that we do binary search on the segment tree
using namespace std;
const int n_bits=19;
const long long inf = 1e18;
long long minseg[1<<(n_bits+1)];
long long maxseg[1<<(n_bits+1)];
long long lazyadd[1<<(n_bits+1)];
// a standard lazy propagation segment tree
// here we need to support both min and max
// so it is essentially 2 segtrees combined together
// but we only need 1 copy of lazy add
struct segtree {
long long last_value = 0;
long long small = inf;
long long big = -inf;
segtree() {}
void update(int node, int change) { // treated as a suffix update
last_value += change;
node += (1<<n_bits);
lazyadd[node] += change;
while(node>1) {
if(node%2==0) {
lazyadd[node+1] += change;
}
minseg[node/2] = min(minseg[node]+lazyadd[node], minseg[node^1]+lazyadd[node^1]);
maxseg[node/2] = max(maxseg[node]+lazyadd[node], maxseg[node^1]+lazyadd[node^1]);
node = node/2;
}
}
int solve(int capacity) { // returns the largest index i, such that the range >= c
int node = 1;
small = inf;
big = -inf;
long long lz = 0;
while(node < (1<<n_bits)) {
lz += lazyadd[node];
node *= 2;
if(max(big, maxseg[node+1]+lazyadd[node+1]+lz) - min(small, minseg[node+1]+lazyadd[node+1]+lz) > capacity) {
node++;
} else {
big = max(big, maxseg[node+1]+lazyadd[node+1]+lz);
small = min(small, minseg[node+1]+lazyadd[node+1]+lz);
}
}
if(minseg[node] + lazyadd[node] + lz < last_value) {
return capacity - (big - last_value);
} else {
return last_value - small;
}
}
};
vector<pair<int,int>> toggle[(int)6e5];
// this tells you what you need to toggle on/off as you move across the boxes
// stores a pair indicating the query id and the change in number of candies
vector<int> distribute_candies(vector<int> C, vector<int> L, vector<int> R, vector<int> V) {
int n = C.size();
int q = L.size();
segtree s;
for(int i=0; i<q; i++) {
toggle[L[i]].push_back(make_pair(i, V[i]));
toggle[R[i]+1].push_back(make_pair(i, -V[i]));
}
vector<int> ans;
ans.resize(n);
for(int i=0; i<n; i++) {
for(pair<int,int> p: toggle[i]) {
s.update(p.first+2, p.second); // store values as if the boxes have infinite capacity
}
if(maxseg[1] - minseg[1] < C[i]) { // easy case: range is small
ans[i] = s.last_value - (minseg[1] + lazyadd[1]);
} else { // we binary search on the segtree
ans[i] = s.solve(C[i]);
}
}
return ans;
}
