c++ multidimensional array?

Question

Hi.I want use two dimens. array for print screen this: 1 6 7
2 5 8
3 4 9.
I tried this:
a[3][3]={1,6,7,2,5,8,3,4,9}

for (i=0;i<3;i++){
for (j=0;j<3;j++){ cout< cout< }
Its give me true but i think its wrong way because:
if i have a[3][3]={1,2,3,4,5,6,7,8,9} how i can print this: 1 6 7
2 5 8
3 4 9.
please write code,i searched google but i cant find.Thanks! Sorry for my bad english.

Answer 1

The correct initialisation should be

a[3][3] = {{1,6,7}, {2,5,8}, {3,4,9}};



Try

int b[3][3];

int k = 0;

for (i=0;i<3;i++){

for (j=0;j<3;j++) {

if (i % 2 == 0) {

b[i][k] = a[i][j];

k++;

} else {

b[i][k] = a[i][j];

k--;

} //End if

}

if (i % 2 == 0) {

k = 2;

} else {

k = 0;

} //End if

}



for (i=0;i<3;i++){

for (j=0;j<3;j++){ cout<
cout<
}



Have fun.

Answer 2

Hi,

You can do one of two things:-

Your input is a[3][3]={1,2,3,4,5,6,7,8,9}

You need output {1,6,7,2,5,8,3,4,9}

1.Assuming your input is always a 2-d array a[3][3]

You can use a temporary variable say temp and rearrange the sequence

Something like

for(n=0 to 8)

{

a[0][0]=a[0][0];//Keeping 1 in the first position

temp=a[0][1];//Storing value in second position say 2

a[0][1]=a[1][2]//Now we are storing the value in position a[1][2] that is 6 in the second position

a[1][2]=temp//At this stage your array will look like this a[3][3]={1,6,3,4,5,2,7,8,9}

//Similarly you assign the value for next position

//Say you want 7 to appear in the 3rd position

/*so add a few steps and exchange array element 7 with array element 3 so you get the array a[3][3]={1,6,7,4,5,2,3,8,9} */

//proceed this way until you get the required arrangement

}



2.The other way is a more complicated code but it is shorter,I suggest you dont attempt it since it looks like you are at a beginners level

Answer 3

As you can see from previous answers, there are a few different ways to approach this. If you find that your code has special-case logic for this particular array or its contents, it's a hint that you're not getting the best solution. As I've done it below, the only restriction is that the matrix is square.



#include



using namespace std;



const size_t rows = 3;

const size_t cols = rows;



int A[rows][cols] = {

    {1, 2, 3},

    {4, 5, 6},

    {7, 8, 9}

};



int main(int argc, char *argv[]) {

    //

    // Display transpose of square matrix A, but with

    // odd-numbered columns reversed, top to bottom.

    //

    for (size_t i = 0; i < rows; i++) {

        for (size_t j = 0; j < cols; j++) {

            cout << A[ j ][ (j & 1) ? (cols - i - 1) : i ] << ' ';

        }

        cout << endl;

    }

    return 0;

}



#if 0



Program output:



1 6 7

2 5 8

3 4 9



#endif

Answer 4

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Answer 5

include

using namespace std;



int main() {

  int a[3][3] = {1,2,3,4,5,6,7,8,9};

  for (int i = 0; i < 3; i++) {

    for (int j = 0; j < 3; j++) {

      if (a[i][j] == 1 || a[i][j] == 6 || a[i][j] == 7)

        cout << a[i][j] << " ";

    }

  }

  return 0;

}