Three methods of generating arbitrary odd order magic squares

1. Magic square Matrix

Magic Square is a very interesting digital matrix, in the early famous nine gossip array on the magic side.

The magic square is defined as:

1 to n*n integers are filled in the squares of the n*n, and the sum of the numbers of each row and column and diagonal must be equal.

You, as the top programmer at the gossip company, now need you to solve a problem and find any odd-numbered magic square.

Input:

 

The input includes multiple test sets, one positive odd n (1<= n < 1000) for each behavior, and 0 as the end of the input and does not need to be processed.

 

Output:

 

For each n input, output one of its corresponding n-order magic Square, if there are multiple, any one can.

Each magic square is a matrix of n*n,

For each magic square, a row of magic squares is output per line, with one or more spaces separating the digits in each row. Different magic squares are separated by a blank line.

Algorithm idea:

The problem is also implemented in three ways.

Main interface:

Generation of odd-order magic squares by Merzirac method

Algorithm idea:

In the first row of the box in the center of 1, and then fill the right Top 2, 3, 4 ..., if there is already a number on the top right, move down one to continue filling.

Input 1: Create a 5*5 Rubik's Cube, this Rubik's Cube is fixed. You can change the n in the macro definition to produce any odd order Rubik's Cube.

The Loubere method generates the odd order magic square.

Algorithm idea:

In the center of the square up in a box 1, in turn, fill 2, 3, 4 ... if there is already a number on the top right, move up two squares and continue filling

Input 2: Create a 5*5 Rubik's Cube, this Rubik's Cube is fixed. You can change the n in the macro definition to produce any odd order Rubik's Cube.

The horse method generates the odd order magic square.

Algorithm idea:

Put 1 in any grid first. Take 1 steps to the left, and go down 2 steps into 2 (called horse stance), walk 1 steps to the left, and go 2 steps into 3, and then put to N. Place the n+1 below N (called a jump), and then put the above method into the 2n, placing the 2n+1 at the bottom of the 2n.

Input 3: Prompt, enter a data, at this time the input data indicates how many different fantasy parties randomly generated, but also can change the definition of the "n" Create fantasy side.

Enter 0: Any key to exit.

Attachment:

#include <stdio.h>

#include <windows.h>

#include <time.h>

void Merziracmagic ();

void Magichorserandom ();//Randomization method

void Louberemagic ();

int num (int number); The number of magic squares produced by Magichorserandom () Randomization method

int Menu_select ();

#define N 5//magic square must be odd

int Menu_select ()

{

char c;

System ("CLS");/* Run the splash screen/*

printf ("\t\t**** Rubik's Cube Author byyongliu ****\n");

printf ("\t\t 1. Merziracmagic Rubik's Cube |\n ");

printf ("\t\t 2. Louberemagic Rubik's Cube |\n ");

printf ("\t\t 3. Magichorserandom Rubik's Cube |\n ");

printf ("\t\t 0. Exit |\n ");

printf ("\t\t*****************************************\n");

printf ("\t\t\t Please Choose (0-3):");

do{

C=getchar (); /* Read the choice * *

}while (c< ' 0 ' | | C> ' 3 ');

Return (c ' 0 '); /* Return selection * *

}

int main ()

{

int number;

for (;;)

{

Switch (Menu_select ())//* Select Judgment * *

{

Case 1:

printf ("\t\t1. Merziracmagic Rubik's Cube |\n ");

Merziracmagic ();

Break

Case 2:

printf ("\t\t2. Louberemagic Rubik's Cube |\n ");

Louberemagic ();

Break

Case 3:

printf ("\t\t3. Magichorserandom Rubik's Cube |\n ");

printf ("\t\tpleaseinput The number of magic");

scanf ("%d", &number);

Num (number);//Randomization method

printf ("\t\t\t");

Break

Case 0:

printf ("\t\t\t End Exit!\n"); /* END PROCEDURE * *

printf ("\t\t\t");

System ("pause");

Exit (0);

}

}

}

int num (int number)

{

inti=0;

intcount=0;

Srand ((int) time (NULL));

while (++count<=number)

{

printf ("\t\t%d.\n", ++i);

Magichorserandom ();

}

RETURN1;

}

void Magichorserandom ()

{

inta[n+1][n+1]={0};//to initialize it

Inti=rand ()%n+1; IJ randomly selected in 1 to N*n

Intj=rand ()%n+1;

printf ("%2d\n", I);

printf ("%2d\n", j);

A[i][j]=1;

for (intk=2;k<=n*n;k++)

{

int row=i;

int col=j;

++j; Take one step to the right

if (j>n)

J=1;

++i; Take a step down

if (i>n)

I=1;

++i; Take a step down

if (i>n)

I=1;

if (a[i][j]==0)//If there is no element in the upper right to fill in

A[i][j]=k;

Else

{

I=row; Remember subscript

J=col;

i++;

if (i>n)

I=1;

A[i][j]=k;

}

}

for (i=1;i<=n;i++)

{

printf ("\t\t");

for (j=1;j<=n;j++)

{

printf ("%3d", A[i][j]);

}

printf ("\ n");

}

printf ("\ n");

printf ("\ n");

}

Generation of odd-order magic squares by Merzirac method

In the first row of the box in the center of 1, and then fill the right Top 2, 3, 4 ..., if there is already a number on the top right, move down one to continue filling.

void Merziracmagic ()

{

inta[n+1][n+1]={0};//to initialize it

inti=1,j= (n+1)/2;

A[i][j]=1; The middle element of the first line is placed 1

for (intk=2;k<=n*n;k++)

{

int row=i;

int col=j;

I.;

++j; Fill in the top right element continuously

if (i==0)

I=n;

if (j>n)

J=1;

if (a[i][j]==0)//If there is no element in the upper right to fill in

A[i][j]=k;

Else

{

I=row; Remember subscript

J=col;

i++;

if (i>n)

I=1;

A[i][j]=k;

}

}

for (i=1;i<=n;i++)

{

printf ("\t\t");

for (j=1;j<=n;j++)

{

printf ("%3d", A[i][j]);

}

printf ("\ n");

}

printf ("\ n");

printf ("\ n");

}

Generation of odd-order magic squares by Loubere method

In the center of the square up in a box 1, in turn, fill in 2, 3, 4 ..., if there is already a number on the top right, move up two squares to continue filling.

void Louberemagic ()

{

inta[n+1][n+1]={0};//to initialize it

Inti= (n+1)/2-1,j= (n+1)/2;

A[i][j]=1; The upper layer of the most intermediate element

for (intk=2;k<=n*n;k++)

{

int row=i;

int col=j;

I.;

++j; Fill in the top right element continuously

if (i==0)

I=n;

if (j>n)

J=1;

if (a[i][j]==0)//If there is no element in the upper right to fill in

A[i][j]=k;

Else

{

I=row; If you use backtracking methods to consider the problem more, or remember subscript

J=col;

i--;

if (i==0)

I=n;

i--;

if (i==0)

I=n;

A[i][j]=k;

}

}

for (i=1;i<=n;i++)

{

printf ("\t\t");

for (j=1;j<=n;j++)

{

printf ("%3d", A[i][j]);

}

printf ("\ n");

}

printf ("\ n");

printf ("\ n");

}

Note: Code uploaded in http://download.csdn.net/detail/liuyongvs2009/7077279