分块 （区间修改，单点查询

题目链接https://loj.ac/problem/6277

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int dir[10][10] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int pos[maxn], add[maxn], a[maxn], r[maxn], l[maxn], n;

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

inline void change(int L, int R, int d)
{
    int p = pos[L], q = pos[R];
    if(p == q) for(int i = L; i <= R; ++i) a[i] += d;//如果在一组里，直接暴力修改
    else
    {
        for(int i = p + 1; i <= q - 1; ++i) add[i] += d;//将i的后一组和j的前一组之间的所有组加d
        for(int i = L; i <= r[p]; ++i) a[i] += d;//暴力修改l到l右边界各值
        for(int i = l[q]; i <= R; ++i) a[i] += d;//暴力修改r左边界到r各值
    }
}
int main()
{
    std::ios::sync_with_stdio(false);std::cin.tie(0);
    n = read();
    for(int i = 1; i <= n; ++i)
    {
        a[i] = read();
    }
    int t = sqrt(n);//总共t块
    for(int i = 1; i <= t; ++i)
    {
        l[i] = (i - 1) * t + 1;
        r[i] = i * t;
    }//定每块左边界，右边界。

    if(r[t] < n) t++, l[t] = r[t - 1] + 1, r[t] = n;//为多出那块赋左右边界值，再让块总数加1.
    for(int i = 1; i <= t; ++i)
    {
        for(int j = l[i]; j <= r[i]; ++j)
        {
            pos[j] = i;
        }
    }//每个数字定块，j数属于i块。

    for(int i = 1; i <= n; ++i)
    {
        int opt, L, R, C;
        opt = read(), L = read(), R = read(), C = read();
        if(!opt) change(L, R, C);
        else
        {
            int q = pos[R];
            cout<<a[R] + add[q]<<'
';//查询q的值
        }
    }
    return 0;
}