【每日基础算法】树状数组 - 动态求连续区间和

for (int i = x; i <= n; i += lowbit(i)) {
   c[i] += k;
}

for (int i = x; i; i -= lowbit(i)) {
    sum += c[i];
}

10 5
1 2 3 4 5 6 7 8 9 10
1 1 5
0 1 3
0 4 8
1 7 5
0 4 8

11
30
35

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int n, m;
int c[N];

//lowbit返回二进制第一个非零的1
int lowbit(int x) {
    return x & -x;
}
//在第x个位置上，加上k
void add(int x, int k) {
    for (int i = x; i <= n; i += lowbit(i)) c[i] += k;
}
//返回x的前缀和
int query(int x) {
    int res = 0;
    for (int i = x; i; i -= lowbit(i)) res += c[i];
    return res;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        int x;
        scanf("%d", &x);
        add(i, x);
    }
    while (m--) {
        int k, a, b;
        scanf("%d %d %d", &k ,&a, &b);
        if (k) add(a, b);
        else printf("%d\n", query(b) - query(a - 1));
    }
    return 0;
}