LOJ - #6279. 数列分块入门 3

题目链接：#6279. 数列分块入门 3

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int blo, n, opt, l, r, c, a[maxn], pos[maxn], atag[maxn];
set<int>st[maxn];
int add(int l, int r, int c)
{
   for (int i = l ; i <= min(pos[l]*blo, r); i++){
    st[pos[l]].erase(a[i]);
    a[i] += c;
    st[pos[l]].insert(a[i]);
   }
   if(pos[l] != pos[r]){
    for (int i = (pos[r]-1)*blo+1; i <= r; i++){
        st[pos[r]].erase(a[i]);
        a[i] += c;
        st[pos[r]].insert(a[i]);
    }
   }
   for (int i = pos[l]+1; i <= pos[r]-1; i++) atag[i] += c;
}
int query(int l , int r, int c)
{
   int ans = -1;
   for (int i = l ; i <= min(pos[l]*blo, r); i++){
    int val = a[i] + atag[pos[l]];
    if(val < c) ans = max(ans, val);
   }
   if(pos[l] != pos[r])
    for(int i = (pos[r]-1)*blo+1; i <= r; i++){
     int val = a[i] + atag[pos[r]];
     if(val < c) ans = max(ans, val);
   }
   for (int i = pos[l] + 1; i <= pos[r]-1; i++) {
    int x = c - atag[i];
    set<int>::iterator it = st[i].lower_bound(x);
    if(it == st[i].begin()) continue;
    --it;
    ans = max(ans, *it+atag[i]);
   }
   return ans;
}
int main()
{
    scanf("%d", &n);
    blo = sqrt(n);
    for (int i = 1; i <= n; i++) scanf("%d",&a[i]), pos[i] = (i - 1) / blo + 1, st[pos[i]].insert(a[i]);

    for (int i = 0; i < n; i++){
        scanf("%d%d%d%d", &opt, &l, &r, &c);
        if(opt == 0){
            add(l, r, c);
        }
        else{
            printf("%d\n", query(l, r, c));
        }
    }
    return 0;
}