鏈接：https://loj.ac/problem/6279

思路：

和上一道一樣的方法，每個塊排序，兩端的塊暴力找，中間的塊二分找到比他小的最大的數，最後取最大的符合條件的數。

實現代碼：

#include<bits/stdc++.h>
using namespace std;
const int M = 1e5+10;
vector<int>v[M];
int a[M],block,n,bl[M],tag[M];
void reset(int x){
    v[x].clear();
    for(int i = (x-1)*block+1;i <= min(x*block,n);i ++)
        v[x].push_back(a[i]);
    sort(v[x].begin(),v[x].end());
}
 
void update(int l,int r,int c){
    for(int i = l;i <= min(bl[l]*block,n);i ++){
        a[i] += c;
    }
    reset(bl[l]);
    if(bl[l] != bl[r]){
        for(int i = (bl[r]-1)*block+1;i <= r;i ++)
            a[i] += c;
        reset(bl[r]);
    }
    for(int i = bl[l]+1;i <= bl[r] - 1
 
;i ++)
        tag[i] += c;
}

int query(int l,int r,int c){
    int ans = -1;
    for(int i = l;i <= min(bl[l]*block,r);i ++)
        if(a[i]+tag[bl[l]] < c) ans = max(ans,a[i]+tag[bl[l]]);
    if(bl[l] != bl[r]){
        for(int i = (bl[r]-1)*block+1;i <= r;i ++)
            if(a[i]+tag[bl[r]] < c) ans = max(ans,a[i]+tag[bl[r]]);
    }
    
 
for(int i = bl[l]+1;i <= bl[r]-1;i ++){
        int x = c - tag[i];
        int y = lower_bound(v[i].begin(),v[i].end(),x) - v[i].begin();
        if(y>=1) ans = max(ans,v[i][y-1]+tag[i]);
    }
    return ans;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int f,l,r,c;
    cin>>n;
    block = sqrt(n);
    for(int i = 1;i <= n;i ++) cin>>a[i];
    for(int i = 1;i <= n;i ++){
        bl[i] = (i-1)/block+1;
        v[bl[i]].push_back(a[i]);
    }
    for(int i = 1;i <= bl[n];i ++)
        sort(v[i].begin(),v[i].end());
    for(int i = 1;i <= n;i ++){
        cin>>f>>l>>r>>c;
        if(f == 0) update(l,r,c);
        else cout<<query(l,r,c)<<endl;
    }
    return 0;
}

Loj 6278 數列分塊入門 3