Tree-like array [data structure]

Tree-like array

——! x^n+y^n=z^n

Well, the picture is online search ...

　　

We make

C[1]=A[1]

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

C[3]=A[3]

C[4]=A[4]+C[3]+C[2]

C[5]=A[5]

C[6]=C[5]+A[6]

C[7]=A[7]

C[8]=A[8]+C[4]+C[6]+C[7]

C[9]=A[9]

In fact, we are the c[n] of the jurisdiction of the I element, and this I is n to binary when the first occurrence from the forward 1 and the next 0 of the size of the number of components (recorded as Lowbit (n)).

Give an example c[6]=a[5]+a[6]

6= (2,lowbit) (6) = (10) 2=2;

But this will produce a problem, how to beg lowbit?

Using the computer's complement feature we can get:

Lowbit (x) =x& (×). [Know the complement of the students hand push can, do not know hurriedly Baidu]

Of course, as a data structure to have his usefulness.

With a tree-like array we can do two simple operations:

<i> the value of a[1]+a[2]+l+a[x]

<ii> Single-point update

<i> Zone Search

Suppose we ask for 1~x interval and we can first query c[x], obviously c[x] is contained in the scope of our request;

However, when we have finished querying c[x], we should continue to query the non-queried parts;

Note: c[x] Jurisdiction: [X-lowbit (x) +1,x], so we should continue to query C[x-lowbit (x)];

Repeat the above operation until the query is complete.

Code:

1 int getsum (x) {2     int res=0; 3      for (; x;x-=lowbit (x)) 4         res+=bit[x]; 5     return Res; 6 }

My tree-like array has always liked to play a bit ah ...

<ii> Single-point update

We hope A[x]+inc

Of course we will let C[x]+inc first;

Then we need to find the c[y] that governs c[x], because x must ∈[y-lowbit (y) +1,y], how do we do this?

In fact, we just need to make a 0 on the back of X, just think about it.

Code:

1 void Update (x,inc) {2      for (; x<=n;x+=lowbit (x)) 3         Bit[x]+=Inc; 4 }

What, do you think the code is a lot shorter? (Compare line tree)

Tree-like array [data structure]