Derivation of formula + Yang Hui's triangle + xor or property + feed representation XOR Matrix hackerrank-xor-matrix__ formula derivation

Do you:
1 Knowledge Points: "Formula derivation + Yang Hui's triangle + xor or character + feed representation"
2:
Define XOR or matrix elements a[i, J]:

The first line enters N, M represents the Xor of N row m columns, and the second row enters n elements to represent the first row of the XOR or matrix, and the title asks us to output the row of the XOR or matrix.
2.1 Data range:
·1 <= N <= 1e5
·1 <= m <= 1e18
·1 <= A[i, j] <= 1e9
3 Problem Solving Process
3.1 Formula derivation:
A[i, j] = A[i-1, j] ^ a[i-1, j+1]
That is, for A[i, J] a single element:
A[0, j] = A[0, j]
A[1, j] = A[0, j] ^ a[0, j+1]
A[2, j] = A[1, j] ^ a[1, j+1] = (a[0, j] ^ a[0, j+1]) ^ (a[0, j+1] ^ a[0, j+2])
= A[0, j] ^ a[0, j+1] ^ a[0, j+1] ^ a[0, j+2]
A[3, j] = A[2, j] ^ a[2, j+1] = (.) ^ (.)
= A[0, j] ^ a[0, j+1] ^ a[0, j+1] ^ a[0, j+1] ^ a[0, j+2] ^ a[0, j+2] ^ a[0, j+2] ^ a[0, j+3]
.
.
.
That
A[i, J]⇦a[0, J], A[0, j+1], a[0, j+2] , A[0, J+i]
And then push it:
A[x+i, J]⇦a[x, J], A[x, j+1], a[x, j+2] , A[x, J+i]

3.2 Yang Hui's triangle
Will A[i, j] by A[0, J], A[0, j+1], a[0, j+2], , A[0, J+i] indicates that the discovery coefficient satisfies the Yang Hui's triangle:

3.3 Different or property
A[i, J] ^ a[i, j] = 0
A[i, j] ^ 0 = a[i, j]
A[i by Xor, J] ^ a[i, j] = 0 and A[i, j] ^ 0 = 0 Degenerate coefficient matrix
The new coefficient matrix is:
0.1
1.1 1
2.1 0 1
3.1 1 1 1
4.1 0 0 0 1
5.1 1 0 0 1 1
6.1 0 1 0 1 0 1
7.1 1 1 1 1 1 1 1
8.1 0 0 0 0 0 0 0 1
9.1 1 0 0 0 0 0 0 1 1
...
That
A[0, j] = A[0, j]
A[1, j] = A[0, j] + a[0, j+1]
A[2, j] = A[0, j] + a[0, j+2]
A[3, j] = A[0, j] + a[0, j+1] + a[0, j+2] + a[0, j+3]
A[4, j] = A[0, j] + a[0, j+4]
A[5, j] = A[0, j] + a[0, j+1] + a[0, j+4] + a[0, j+5]
A[6, j] = A[0, j] + a[0, j+2] + a[0, j+4] + a[0, j+6]
A[7, j] = A[0, j] + a[0, j+1] + a[0, j+2] + a[0, j+3] + a[0, j+4] + a[0, j+5] + a[0, j+6] + a[0, j+7]
A[8, j] = A[0, j] + a[0, j+8]
A[9, j] = A[0, j] + a[0, j+1] + a[0, j+8] + a[0, j+9]
It can be found that for a row with a power of 2, only through A[0, J] and A[0, J+i] Two two-factor coefficients can be computed a[i, j], namely:

3.4 Binary means:

4. "Recommended Reference" code:

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

#define N 111111
int a[n];
int b[n];
int n;

void Move (int k) {
    //Compute the 2^k ' th row for
    (int i = 0; i < n; i++) b[i] = A[i] ^ a[(i + (1LL << k)) % n];

    Copy back to ' a ' for
    (int i = 0; i < n; i++) a[i] = B[i];
}
int main () {
    //Take the input
    long long m;
    scanf ("%d%lld", &n, &m);
    for (int i = 0; i < n; i++) scanf ("%lld", &a[i));

    Compute the answers
    m--;
    for (int k = 0; k < a k++) {
        if (M & (1LL << k)) move (k);
    }

    Print the answers
    for (int i = 0; i < n; i++) printf ("%d", A[i]);

Vjudge Topic link

The following is the accepted code

 #include <bits/stdc++.h> using namespace std;
typedef long Long LL;

const int N = 104014;

int n, A[n], b[n];

void Move (int k);
  int main () {LL m;
    while (~scanf ("%d%lld", &n, &m)) {for (int i = 0; i < n; i++) scanf ("%d", &a[i));
    m--;
    for (int k = 0; k </k++) {if (M & (1LL << k)) move (k); for (int i = 0; i < n; i++) printf ("%d%c", a[i], i = = n-1?
  ' \ n ': ');
return 0;

  void Move (int k) {for (int i = 0; i < n; i++) B[i] = a[i]^ (a[(i+ (1ll<<k))%n]);
  for (int i = 0; i < n; i++) a[i] = B[i];
Return }