High-dimensional prefixes and

We often use prefixes and.

One-dimensional:

 for (int i=1; i<=n;i++) b[i]=b[i-1]+a[i];

Two-dimensional:

 for (int i=1; i<=n;i++)      for (int j=1; j<=m;j++)         b[i][j]=b[i-1][j]+b[i][j-1]-b[i- 1] [j1]+a[i][j];

What if it's three-dimensional?

 for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++)         for(intk=1; k<=p;k++) B[i][j][k]=b[i-1][j][k]+b[i][j-1][k]+b[i][j][k-1]                            -b[i-1][j-1][k]-b[i-1][j][k-1]-b[i][j-1][j-1]                            +b[i-1][j-1][k-1]

is actually a tolerant.

However, as the dimension T becomes higher, the complexity of the repulsion is 2^t, and the total complexity O (n^t*2^t cannot withstand.

We also have one way:

One-dimensional:

 for (int i=1; i<=n;i++)     a[i]+=a[i-1];

Two-dimensional:

 for (int i=1; i<=n;i++)      for (int j=1; j<=m;j++)        a[i][j]+=a[i][j-1];  for (int i=1; i<=n;i++)      for (int j=1; j<=m;j++)        a[i][j]+=a[i-1][j];

This means that the first pass prefix and, each position a[i][j] is, I line before J's and.

The second time, all the lines in front and added.

Two times to achieve the goal. Looks like trouble.

What about the three dimensions?

 for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++)         for(intp=1;p <=k;p++) A[i][j][k]+=a[i-1][j][k]; for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++)         for(intp=1;p <=k;p++) A[i][j][k]+=a[i][j-1][k]; for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++)         for(intp=1;p <=k;p++) A[i][j][k]+=a[i][j][k-1];

In fact, the two-dimensional understanding is the same. Once again, add the third dimension and the past.

However, this three-dimensional only 3 times, that is, for t-dimensional, in fact, as long as O (n^t*t) complexity is very low.

In fact, we actually solve the problem, often use is the situation of n=2.

Like what

Examples:

1. Section and (Cattle network Noip pre-match Camp-popularization Group (fourth))

Enter an array of length n A[i], subscript starting from 0 (0 to n-1)
ensure that N is a power of 2 for the entire number of times,
for each i (0 <= i < n)
all satisfies ((I & j) = = j) The sum of a[j]. n<=2^20that is, the subset of each I and. If n=2^6, if the binary representation of I: 10101 as a 5-dimensional coordinate,so, the subset of I is the high-dimensional prefix of this coordinate and. It can be found that the N of each dimension is 2,this is better handled. if it is General: W represents the highest dimension:

 for (int i=0; i<w;i++) {    for (int j=0;j< (1< <W) (j + +){        if(j& (1<<i)) f[j]+=f[j^ (1<<W)] ;     }}

Beside

#include <bits/stdc++.h>using namespaceStd;typedefLong Longll;Const intN= (1<< +); ll A[n];intN;intMain () {scanf ("%d", &n);intp=0;  for(intI=0; i<n;i++) scanf ("%lld",&A[i]);  for(intI=1; i<n;i<<=1) {p++;  for(intj=0; j<n;j++){            if((j& (1<<p-1))) a[j]+=a[(j^ (1<<p-1))]; }    } for(intI=0; i<n;i++) printf ("%lld\n", A[i]);return 0;}

complexity and high-dimensional prefixes are the same; O (2^t*t)

High-dimensional prefixes and