LOJ block entry columns 1

LOJ block entry columns 1

topic:

• link

answer:

• Addition to each block provided a flag (this block is to record the number of elements added together), each operation (1) marking each block O directly, without the entire block is relatively small due to the elements, the elements modifying violence value.

#include <iostream>
#include <cstdio>
#include <cmath>
#define N 50005
using namespace std;

struct Block {int l, r, tag;} block[N];
int n, size, num;
int a[N], bel[N];

int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return x;
}

void build()
{
    size = (int)sqrt(n), num = n / size;
    if(n % size != 0) num++;
    for(int i = 1; i <= num; i++)
        block[i].l = (i - 1) * size + 1,
        block[i].r = i * size;
    for(int i = 1; i <= num; i++)
        for(int j = block[i].l; j <= block[i].r; j++)
            bel[j] = i;
}

void update(int l, int r, int w)
{
    if(bel[l] == bel[r])
    {
        for(int i = l; i <= r; i++) a[i] += w;
        return;
    }
    for(int i = l; i <= block[bel[l]].r; i++) a[i] += w;
    for(int i = r; i >= block[bel[r]].l; i--) a[i] += w;
    for(int i = bel[l] + 1; i <= bel[r] - 1; i++) block[i].tag += w;
}

int ask(int pos) {
    return a[pos] + block[bel[pos]].tag;
}

int main()
{
    cin >> n;
    for(int i = 1; i <= n; i++) a[i] = read();
    build();
    for(int i = 1; i <= n; i++)
    {
        int op = read(), l = read(), r = read(), w = read();
        if(!op) update(l, r, w);
        else if(op == 1) printf("%d\n", ask(r));
    }
    return 0;
}