"Tree-like array"

Question: Can we use linear data structure to solve the problem of dynamic statistics subtree right ? Yes, a tree-like array.

Assuming the current array is a[], the number of elements is n.

1. The weights and arrays of the sub-interval are sum, then the array elements of the interval between I and J in the array a[] and the summation of sum[i,j]= a[k] "K Belongs to (I->J)"

2. Now define the prefix and array s[],s[i] to represent the sum from a[i]---a[i], then this can be represented sum[i,j] = s[j]-s[i-1].

3. Lowbit (k) is the number represented by the first 1 on the right of the binary representation of the integer k, lowbit (k) =k& (-K).

4. The tree array for c[],c[k] stores the sum of the lowbit (k) elements from the a[k] beginning to the lower subscript, and the first layer of traversal.

Note: We want to store the elements of the a[] array starting with subscript 1.

Here's a list:

C[1]=A[1]; S[1]=C[1];

C[2]=A[2]+A[1]; S[2]=C[2];

C[3]=A[3]; S[3]=C[3]+C[2];

C[4]=A[4]+A[3]+A[2]+A[1]; S[4]=C[4];

C[5]=A[5]; S[5]=C[5]+C[4];

C[6]=A[6]+A[5]; S[6]=C[6]+C[4];

C[7]=A[7]; S[7]=C[7]+C[6]+C[4];

C[8]=A[8]+A[7]+A[6]+A[5]a[4]+a[3]+a[2]+a[1]; S[8]=C[8];

C[9]=A[9]; S[9]=C[9]+C[8];

5. The complexity of calculating each s[i] is O (log2 (n)).

6. Code:

#include <stdio.h> #include <string.h>int a[101];int c[101];int s[101];int lowbit (int x) {    return x& (-X);} int sum (int x) {    int s=0;    while (x>0) {        s+=c[x];        X=x-lowbit (x);    }    return s;} int main () {    int i, j, K;    for (I=1; i<=10; i++)        a[i]=i;//create a[] array    for (i=1; i<=10; i++) {//calculate each c[i]        C[i]=0;//c[i] from initial 0        For (J=i-lowbit (i) +1; j<=i; J + +) {            c[i]+=a[i];}    }    for (I=1; i<=10; i++) {        s[i]=sum (i);//calculate the prefix array for I and    }    return 0;}

"Tree-like array"