树状数组Binary Indexed Tree(2):区间修改，单点查询

Description

给定数列a[1],a[2]…,a[n],需要依次执行q个操作。
操作分两类：
1 l r x : 给定l，r，x，对于所有 i ∈ \in ∈ [l,r]，将a[i]加上x
2 i ：求a[i]的值
(1 ≤ \leq ≤ n,q ≤ \leq ≤ 10⁶)

解题思路

算法标签：树状数组
利用差分的思想。设原数组number[i],d[i] = number[i] - number[i-1]，所以number[i] = ∑ j = 1 i d [ j ] \sum_{j=1}^id[j] ∑j=1i​d[j]，即可以通过求d[i]的前缀和查询。修改通过给d[l]加上x，给d[r+1]减上x

代码

// TSWorld
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1000009;

long long tree[N];
int number[N];
int n, q;

inline int lowbit(long long x) { return x & (-x); }
void Change(int index, int delta) {
    while (index <= n) {
        tree[index] += delta;
        index += lowbit(index);
    }
}

long long Query(long long index) {
    long long sum = 0;

    while (index > 0) {
        sum += tree[index];
        index -= lowbit(index);
    }
    return sum;
}
int main() {
    int cmd = 0, l = 0, r = 0, x = 0;
    cin >> n >> q;

    for (int i = 1; i <= n; i++) {
        scanf("%d", &number[i]);
        Change(i, number[i] - number[i - 1]);
    }

    for (int i = 1; i <= q; i++) {
        scanf("%d", &cmd);

        if (cmd == 1) {
            scanf("%d%d%d", &l, &r, &x);
            Change(l, x);
            Change(r + 1, -x);
        } else {
            scanf("%d", &l);
            printf("%lld\n", Query(l));
        }
    }

    return 0;
}