# C + + implementation: The spiral Matrix Instance code _c language

Source: Internet
Author: User
Tags int size

It is observed that the subscript of the matrix has such a rule: a row increments after column B increases then the C line decreases and then the D column decreases, but the corresponding value is gradually increased. So there are 4 loops to implement, and it's worth noting that you don't overwrite previous values when assigning values. So here you choose the same color section assignment, the code is as follows:

Copy Code code as follows:

#include <iostream>
#include <iomanip>

using namespace Std;

Output Helix matrix
void Matrix ()
{
const int size = 10; Matrix size
int Matrix[size][size] = {0};

int row = 0;
int col = 0;

int start = 1;//start value
int temp = size;
for (int count = 0; count < SIZE/2 count++)//size order matrices can draw Size/2 loops
{
&nb sp;       for (; col < temp-1; col++)//A row assignment
Matrix[row][col] = start++;
for (; row < temp-1; row++)//b row Assignment
Matrix[row][col] = start++;
for (col = temp-1 col > count; col--)//C row Assignment
&n bsp;        Matrix[row][col] = start++;
for (row = temp-1 row > count; row--)//D row Assignment
&n bsp;        Matrix[row][col] = start++;

Go to the next lap
temp--;
row++;
start = 1; This is 1 because it's going to be 1 more when you change the loop.
}

if (0!= size% 2)///if size is odd then there will be a number of the last one to traverse.
Matrix[row][col+1] = start + 1;

Output array
for (int i = 0; i < size; i++)
{
for (int j = 0; J < size; J + +)
{
cout << SETW (5) << matrix[i][j];
}
cout << Endl;
}
}

int main (int argc, char **argv)
{
Matrix ();
return 0;
}

The results are as follows (odd, even):

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