Data Structure - Fenwick tree

Article Directory

1. Fenwick tree
2. Single point query, modify a single point
3. Single point query, modify the interval

1. Fenwick tree

Similar data structure: segment tree (Segment Tree)

Fenwick tree with the tree line difference:

Fenwick tree can do, segment tree can do! (Tree line features more cattle)
Fenwick tree code is simple, easy to implement than the segment tree (tree code simpler array)

Query and modify complex array of tree are $\log(n)$

Here Insert Picture Description

A former array is
Fenwick tree is C

C1 = A1
C2 = C1 + A2 = A1 + A2
C3 = A3
C4 = C2 + C3 + A4 = A1 + A2 + A3 + A4
C5 = A5
C6 = C5 + A6 = A5 + A6
C7 = A7
C8 = C4 + C6 + C7 + A8 = A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8

To C8 (2 binary 8 is 1000, there are three behind 0, so he was in possession with the number 3 C4、C6、C7, as well as their own A8)
No end of odd 0, so odd bit only their own
Then how to calculate the number of tubes with a few numbers it?

2. Single point query, modify a single point

Tree array core function lowbit(int m): the role of obtaining a position m 1 binary representation of the end, to query for the prefix and m, m = m - lowbit(m)on behalf of the end of the continuous binary 1 1 operation continuously performed until m == 0 completed, can be obtained which is composed of a prefix and a few Cm

int lowbit(int m){
    return m & (-m);//(获取最后一个1的10进制值)
}
int getsum(int i){    //求A[1],A[2],...A[i]的和
    int res = 0;
    while(i > 0){
        res += c[i];
        i -= lowbit(i);
    }
    return res;
}

Eight ( 1000) to count follows:
NOTE: negative binary numbers are positive end negated +1

8 & (-8) = 1000 & (.111..1000) = 1000 = 8....8-8=0,sum=C8=A[1]+..A[8]
-------------
6 & (-6) =  0110 & (1010) = 10 = 2 .... 6-2=4,sum=C6
4 & (-4) = 0100 & (1100) = 100 = 4.....4-4=0,sum+=C4=C6+C4=A[1]+...A[6]
-------------
奇数上面操作结果均为 1

#include <bits/stdc++.h>
using namespace std;

int n = 8;
int a[9]={-1,1,2,3,4,5,6,7,8}, c[9] = {0}; //对应原数组和树状数组

int lowbit(int x){
    return x&(-x);
}

void update(int i, int delta){    //在i位置加上delta
    while(i <= n){
        c[i] += delta;
        i += lowbit(i);
    }
}

int getsum(int i){    //求A[1],A[2],...A[i]的和
    int res = 0;
    while(i > 0){
        res += c[i];
        i -= lowbit(i);
    }
    return res;
}

int main(){
    for(int i = 1; i < 9; ++i)
        update(i,a[i]);//读取原数据，插入树状数组
    cout << getsum(3) << endl;//获取前3个数的和
    cout << getsum(8) << endl;
    update(3,2);
    cout << getsum(3) << endl;
    cout << getsum(8) << endl;
    cout << getsum(4)-getsum(2) << endl;//获取A[3],A[4]的区间和
    return 0;
}