"Explanations" LibreOJ6277 the number of columns block entry 1

Portal

Description

A given length \ (n-\) is the number of columns, and \ (n-\) operations, the operation involving the addition interval, a single check-point value.

Input

The first input line of a digital \ (n-\) .

The second row input \ (n-\) digits, the \ (I \) digits of \ (a_i \) , separated by a space.

Now enter \ (n-\) line of inquiry, four digits per line \ (opt \) , \ (L \) , \ (R & lt \) , \ (C \) , separated by a space.

If \ (\ texttt opt = {0} \) , represents located \ ([l, r] \ ) numbers are added between \ (C \) .

If \ (\ texttt opt = {}. 1 \) , indicating an inquiry \ (a_i \) values ( \ (L \) and \ (C \) is ignored).

Output

For each inquiry, the output line number indicates the answer.

Sample Input

Output

2
5

Hint

For \ (100 \% \) data, \ (. 1 \ n-Le \ 50000 Le, 31 is -2 ^ {} \ Le Others, ANS \ Le 31 is {2} ^ -. 1 \) .

Solution

Block, the first sequence into \ (\ sqrt {n} \ ) blocks, when the addition interval, the left and right trim piece violent process, the entire block lazy updating flag. Single-point evaluation, as long as the value of their own in which it blocks labeled lazy add it.

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

const int MAXN = 50005;
int n, a[MAXN], bl[MAXN], tag[MAXN];
int main() {
    scanf("%d", &n);
    int block = (int)sqrt(n);//总块数
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        bl[i] = (i - 1) / block;//分块
    }
    for (int i = 1; i <= n; i++) {
        int opt, x, y, val;
        scanf("%d%d%d%d", &opt, &x, &y, &val);
        if (opt == 0) {
            if (bl[x] == bl[y]) {
                for (int i = x; i <= y; i++)
                    a[i] += val;//如果区间不包含任何块
            } else {
                for (; bl[x] == bl[x - 1]; x++)
                    a[x] += val;//处理边角料
                for (; bl[y] == bl[y + 1]; y--)
                    a[y] += val;//处理边角料
                for (int i = x; i <= y; i += block)
                    tag[bl[i]] += val;//更新整块的懒标记
            }
        } else printf("%d\n", a[y] + tag[bl[y]]);//单点求值
    }
    return 0;
}

Attachment

Test data Download: https://www.lanzous.com/i51c8of