Solved here is the code that worked for me
for (i = 0; i < nbLine; i++)
for (k = 0; k < zoom; k++)
{
for (j = 0; j < nbColumn; j++)
for (l = 0; l < zoom; l++)
{
printf("%d ", *(array+ (i*nbColumn) + j));
}
printf("\n");
}
I have a function that output a two dimensional array ([column][line]) and it need to be zoom, in fact it is like going from
1 2
3 4
to a
1 1 2 2
3 3 4 4 array when zoom is equal to 2
here is some code :
static void output(int* array, int nbColumn, int nbLine, int zoom)
{
int i, j, k, l;
for (i = 0; i < nbColumn; i++)
for (k = 0; k < zoom; k++)
{
for (j = 0; j < nbLine; j++)
for (l = 0; l < zoom; l++)
{
printf("%d ", *(array+ (i*nbColomn) + j));
}
printf("\n");
}
}
when I try this code on a squared array, it works fine, but when using a rectangular one, it fails. I have tried to debug it by replacing the printf("%d ", *(array + (i*nbColumn) + j)); by printf("%d ", (i*nbColumn) + j); and it gives me this result for a 8 colomns by 5 rows array :
0 1 2 3 4 5 6 7
5 6 7 8 9 10 11 12
10 11 12 13 14 15 16 17
15 16 17 18 19 20 21 22
20 21 22 23 24 25 26 27
Thanks for help
A working program source is:
#include <iostream>
#include <fstream>
using namespace std;
ifstream in("in.in");
ofstream out("out.out");
int main(void){
int m[100][100],i,j,l,noLines,noColumns,zoom;
in>>noLines>>noColumns>>zoom;
for(i=0;i<noLines;i++){
for(j=0;j<noColumns;j++){
in>>m[i][j];
}
}
for (i = 0; i < noLines; i++){
for (j = 0; j < noColumns; j++){
for (l = 0; l < zoom; l++) {
out<<m[i][j]<<' ';
}
}
out<<'\n';
}
return 0;
}
On the input:
4 3 2
1 2 3 4
5 6 7 8
1 2 3 4
5 6 7 8
you get the output:
1 1 2 2 3 3
4 4 5 5 6 6
7 7 8 8 1 1
2 2 3 3 4 4
Changing the parameters on the first line of the input (i.e. the bidimensional array width, height and the zoom), changes the output.
From your example I can see that you are zooming only horizontally, since the elements are duplicated horizontally, but the number of rows is left intact. So you do not need vertical zoom and therefore your cycle with k is unuseful. This should work:
static void output(int* array, int nbColumn, int nbLine, int zoom)
{
int i, j, l;
for (i = 0; i < nbLine; i++)
{
for (j = 0; j < nbColumn; j++)
for (l = 0; l < zoom; l++)
{
printf("%d ", *(array+ (i*nbColomn) + j));
}
printf("\n");
}
}
Related
I am a beginner in programming, I studied all about C, I started to solve problems from hackerrank.com, there I faced a problem to print a pattern like given below
(the output of problem program):
4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
the input will be an integer which will provide the data for the length of pattern square, here it is 4 in image,
I tried a lot to type a proper logic and I end up with this useless code bellow:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
int n;
scanf("%d", &n);
int array[n- 1 + n][n - 1 + n];
array[(n - 1 + n) / 2][(n - 1 + n) / 2] = 1;
int f[n];
for(int i = 0; i < n; i++) {
f[i] = i;
}
for(int i = 0; i < n - 1 + n; i++) {
for(int j = 0; j < n - 1 + n; j++) {
array[j][i] = n - f[j]; //top
array[i][j] = n - f[j]; //left
array[(2 * n - 1 - 1) - i][j] = n - f[i]; //bottem
array[j][(2 * n - 1 - 1) - i] = n - f[i]; //rigth
}
}
for(int i = 0; i < n - 1 + n; i++) {
for(int j = 0; j < n - 1 + n; j++) {
printf("%d ", array[i][j]);
}
printf("\n");
}
return 0;
}
my logic was to make all four borders correct in for loop which will end at center, but its not working, I want a new logic or to improve my logic, if you want to help me out then please give me the way to solve problem instead of giving me a direct code.
It is observable that the pattern consists of n stacked squares:
Square #0 is drawn with ns.
Square #1 is drawn with n-1s.
...
Square #n-1 is drawn with 1s.
Implementing the above:
void draw_pattern(const size_t n)
{
const size_t dim = 2*n-1;
int array[dim][dim];
for (size_t i = 0; i < n; ++i) { // Outer square #i
// Row #0 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[i][j] = n-i;
// Row #n-1 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[dim-i-1][j] = n-i;
// Col #0 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[j][i] = n-i;
// Col #n-1 of the outer square #i
for (size_t j = i; j < dim-i; ++j)
array[j][dim-i-1] = n-i;
}
print_array(dim, array);
}
This is print_array():
void print_array(const size_t dim, int array[dim][dim])
{
for (size_t i = 0; i < dim; ++i) {
for(size_t j = 0; j < dim; ++j)
printf("%d ", array[i][j]);
printf("\n");
}
}
Output:
4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4
The worst case time complexity is O(n2).
When you get a problem like this, try to dumb it down as much a possible. This square can be separated into 8 same, just rotated "slices" that look like:
4 | 4444 | 4444 | 4
43 | 333 | 333 | 34
432 | 22 | 22 | 234
4321 | 1 | 1 | 1234
... and the same for the bottom half, just flipped.
You can see this in the code bellow, to check what line is writing what part of the square, comment it and you will see what section shows zeroes.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n;
scanf("%d", &n);
int array[2 * n - 1][2 * n - 1];
for(int i = 0; i < 2 * n - 1; i++){
for(int j = 0; j < 2 * n - 1; j++){
array[i][j] = 0;
}
}
int f[n];
for(int i = 0; i < n; i++)
{
f[i] = i;
}
for(int i = 0; i < n; i++)
{
for(int j = i; j < n; j++)
{
array[i][j] = n - i;
array[j][i] = n - i;//top left
array[j][2*n - i - 2] = n - i;
array[i][2*n - j - 2] = n - i;//bottom left
array[2*n - j - 2][i] = n - i;
array[2*n - i - 2][j] = n - i;//top right
array[2*n - i - 2][2*n - j - 2] = n - i;
array[2*n - j - 2][2*n - i - 2] = n - i;//bottom right
}
}
for(int i = 0; i < n - 1 + n; i++)
{
for(int j = 0; j < n - 1 + n; j++)
{
printf("%d ", array[i][j]);
}
printf("\n");
}
return 0;
}
I wanted to know if I could get some help creating a square matrix. I know how to create the matrix, but I need to populate the matrix. I am having trouble following the logic to create the following matrix:
0 1 2 3 4 5
1 2 3 4 5 6
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
Here is what I have in my code thus far:
EDIT*
I have changed the else statement from arr[i][j] += 1 to arr[i][j-1] + 1
void computeMatrix(int rows, int cols, int sqMatrix[][cols]){
int i,j;
for(i = 0; i < rows; i++){
for(j = 0; j < cols; j++){
if(i == 0 && j == 0){
arr[i][j] = 0;
}
else{
arr[i][j] = arr[i][j-1] + 1; //previously arr[i][j] += 1
}
}
}
}
The issue I am having is that this code makes the following matrix:
0 1 2 3 4 5
6 7 8 9 10 11
12 13 14 15 16 17
18 19 20 21 22 23
24 25 26 27 28 29
30 31 32 33 34 35
I am not sure what type of logic to implement to get the matrix correct. I have a general idea, but I am not sure how to implement it to the code. I know that as we go down the rows, we increase by 1, and as we go through the columns, we increase by 1 as well.
Any help I could get will be greatly appreciated!
The elements of the first matrix can be set with:
a[i][j] = i + j;
There's no need for an if statement inside the loop. The whole thing looks like:
void computeMatrix(int rows, int cols, int sqMatrix[][cols])
{
int i,j;
for (i = 0; i < rows; i++) {
for (j = 0; j < cols; j++) {
arr[i][j] = i + j;
}
}
}
void computeMatrix(const size_t rows, const size_t cols, int *arr)
{
for(size_t r = 0; r < rows; r++)
{
for(size_t c = 0; c < cols; c++)
{
arr[r * cols + c] = c + r;
}
}
}
void printmatriz(const int rows, const int cols, int *arr)
{
for(size_t r = 0; r < rows; r++)
{
for(size_t c = 0; c < cols; c++)
{
printf("%2d ", arr[r * cols + c]);
}
printf("\n");
}
}
#define X 6
#define Y 6
int main(void)
{
int arr[X][Y];
computeMatrix(X,Y,&arr[0][0]);
printmatriz(X,Y, &arr[0][0]);
}
How do I create a square matrix in the C programming language?
You think of it as an abstract data type, using C dynamic memory allocation.
This answer gives details and code and uses flexible array members.
Read the Modern C book
I wrote a function to print the below pattern.
For example, if the n value is 4 the pattern is
1
2 7
3 6 8
4 5 9 10
Or if the value of n is 5, then the pattern is
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
My function gives me the first two block but not the next block. I'm stuck here for long time!
My function is
int printPattern(int n) {
int row, column, fwdCtr = 1, evenCtr = 0, ctr = n;
for(row = 1; row <= n; row++) {
fwdCtr = row;
for(column = 1; column <= row; column++) {
if(column % 2 != 0) {
printf("%d ", fwdCtr++);
} else {
evenCtr = fwdCtr + ctr;
printf("%d ", evenCtr);
ctr = ctr - 2;
}
}
printf("\n");
}
}
What I get is
1
2 7
3 6 4
4 5 5 4
Please give suggestions of changes!
The following code should do it:
#include <stdio.h>
void f(int n)
{
for (int i = 0; i < n; ++i)
{
for (int j=0; j<=i; ++j)
{
// Calculate the numbers used so far by previous columns
int x = 0;
for(int v=0; v<j;++v)
{
x = x + (n-v);
}
if ((j % 2) == 0)
{
// even columns
printf("%d ", x+i-j+1);
}
else
{
// odd columns
printf("%d ", x+n-i);
}
}
printf("\n");
}
}
int main(void)
{
f(5);
return 0;
}
Output:
1
2 9
3 8 10
4 7 11 14
5 6 12 13 15
The easy thing to do is just print the right number based on the row and column and the value of n, like this
int main(void)
{
int n = 20;
for (int row = 0; row < n; row++) {
for (int col = 0; col <= row; col++)
printf("%3d ", 1 + col*n - (col-1)*col/2 + (col%2 ? n-1-row : row-col));
printf("\n");
}
}
This is the code that I've tried which simply generates numbers and prints. I am totally stuck about how to access the row numbers and interchange the printing positions of the rows of the matrix.
#include <stdio.h>
int main(void)
{
int i,a[10][10],j,n,count=1;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
a[i][j]=count;
printf("%d\t",count++);
}
printf("\n");
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%d*\t",a[i][j]);
printf("\n");
}
return 0;
}
I am providing a link for the pattern which is to be printed, please check.
https://drive.google.com/open?id=1DKwW8dQggzNjjtAxwPTEI3nRS9AmpK-2Zw
My approach is to avoid the matrix altogether:
#include <stdio.h>
int main() {
int number;
(void) scanf("%d", &number);
int twice = 2 * number;
int squared = number * number;
for (int row = 0, upward = number, downward = 2 * squared; row < number; row++) {
int n = ((upward > squared) ? downward : upward) - number + 1;
for (int column = 0; column < number; column++) {
printf("%d*\t", n++);
}
printf("\n");
upward += twice;
downward -= twice;
}
return 0;
}
EXAMPLES
> ./a.out
3
1* 2* 3*
7* 8* 9*
4* 5* 6*
> ./a.out
4
1* 2* 3* 4*
9* 10* 11* 12*
13* 14* 15* 16*
5* 6* 7* 8*
> ./a.out
5
1* 2* 3* 4* 5*
11* 12* 13* 14* 15*
21* 22* 23* 24* 25*
16* 17* 18* 19* 20*
6* 7* 8* 9* 10*
>
This is my solution to the problem:
#include <stdio.h>
int main() {
int input = 0;
int i = 0;
int j = 0;
int k = 1;
scanf("%d", &input);
int array[input][input];
for(i = 0; i < (input/2); i++) {
for(j = 0; j < input; j++) {
array[i][j] = k;
k++;
}
for(j = 0; j < input; j++) {
array[input-1-i][j] = k;
k++;
}
}
if((input % 2) == 1) {
for(j = 0; j < input; j++) {
array[input/2][j] = k;
k++;
}
}
for(i = 0; i < input; i++) {
for(j = 0; j < input; j++) {
printf("\t%d", array[i][j]);
}
printf("\n");
}
return 0;
}
These are the outputs of the program for diferent inputs:
1
1
2
1 2
3 4
3
1 2 3
7 8 9
4 5 6
4
1 2 3 4
9 10 11 12
13 14 15 16
5 6 7 8
5
1 2 3 4 5
11 12 13 14 15
21 22 23 24 25
16 17 18 19 20
6 7 8 9 10
I'm trying to create a 2d-array in bubble sort, arranged 25 numbers 5 by 5 in ascending order
my inputs
Enter 25 integers:
Input No.[0][0]: 4
Input No.[0][1]: 5
Input No.[0][2]: 8
Input No.[0][3]: 9
Input No.[0][4]: 4
Input No.[1][0]: 2
Input No.[1][1]: 1
Input No.[1][2]: 0
Input No.[1][3]: 2
Input No.[1][4]: 4
Input No.[2][0]: 6
Input No.[2][1]: 7
Input No.[2][2]: 4
Input No.[2][3]: 5
Input No.[2][4]: 5
Input No.[3][0]: 4
Input No.[3][1]: 8
Input No.[3][2]: 9
Input No.[3][3]: 1
Input No.[3][4]: 2
Input No.[4][0]: 4
Input No.[4][1]: 5
Input No.[4][2]: 2
Input No.[4][3]: 1
Input No.[4][4]: 9
my output shows
Ascending:
4 4 5 8 9
0 1 2 2 4
4 5 5 6 7
1 2 4 8 9
1 2 4 5 9
as you can see its not in proper arranged, it only arranged the 5 numbers each lines not the whole numbers
can anybody help arranged my integers like this
Ascending:
0 1 1 1 2
2 2 2 4 4
4 4 4 4 5
5 5 5 6 7
8 8 9 9 9
this is my code so far
int main(){
int rows = 5, cols = 5;
int arr[rows][cols];
int i,j,k,swap;
printf("Enter 25 integers:\n");
for(i = 0; i < rows; i++){
for(j = 0; j < cols; j++){
printf("Input No.[%d][%d]: ", i+0,j+0);
scanf("%d", &arr[i][j]);
}
}
for(k = 0; k < rows; k++){
for(i = 0 ; i < cols; i++){
for(j = i + 1; j < cols; j++){
if(arr[k][i] > arr[k][j]){
swap = arr[k][i];
arr[k][i] = arr[k][j];
arr[k][j] = swap;
}
}
}
}
printf("Ascending:\n");
for( i = 0 ; i < rows; i++){
for( j = 0 ; j < cols; j++){
printf("%3d", arr[i][j]);
}
printf("\n");
}
getch();
}
Improving on Ahmad's answer, I would like to add the following code (for shorting the table in ascending order):
#include <stdio.h>
#define COL 5
#define ROW 6
int main()
{
int temp, t, i, j;
int arr[ROW][COL]={30,29,28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1};
for(t=1; t<(ROW*COL); t++)
{
for(i=0; i<ROW; i++)
{
for(j=0; j<COL-1; j++)
{
if (arr[i][j]>arr[i][j+1])
{
temp=arr[i][j];
arr[i][j]=arr[i][j+1];
arr[i][j+1]=temp;
}
}
}
for(i=0; i<ROW-1; i++)
{
if (arr[i][COL-1]>arr[i+1][0])
{
temp=arr[i][COL-1];
arr[i][COL-1]=arr[i+1][0];
arr[i+1][0]=temp;
}
}
}
for(i=0; i<ROW; i++)
{
printf("\n");
for(j=0; j<COL; j++)
printf("%3d", arr[i][j]);
}
return 0;
}
replace the input with your table and the definitions with the size of your given array and you're done.
output of the above when executed:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
26 27 28 29 30
void twoDimBubbleSort(int** arr, int row, int col) {
for (int i = 0; i < (row * col); ++i) {
for (int j = 0; j < (row * col) - 1; ++j) {
int cr = j / col; // current row
int cc = j % col; // current column
int nr = (j + 1) / col; // next item row
int nc = (j + 1) % col; // next item column
if (arr[cr][cc] > arr[nr][nc])
swap(&arr[cr][cc], &arr[nr][nc]); // any way you want to swap variables
}
}
}
You don't necessarily need to create a 1D array, you can consider your 2D array is a 1D array and transform coordinates when you set/get them.
Consider a structure point with x and y, and ARR_LEN is 5.
int from2Dto1D(point p){ return p.x+ p.y*ARR_LEN;}
Point from1Dto2D(int i){ Point p; p.x = i%ARR_LEN; p.y=i/ARR_LEN; return p;}
Now you can use the normal bubble sorting algorithm with a 1D index on 2D squares array, you just need to convert your index into 2 Point and access/switch data using these Point. (2 because you need a Point with index and a Point with index+1
Put all the array elements from 2-D array to 1-D array then
sort that 1-D array and then put 1-D array in the matrix format
Try this code ....works according to the above given logic
#include<stdio.h>
int main(){
int arr[5][5],l=0;
int result[25],k=0,i,j,temp;
arr[0][0]= 4;
arr[0][1]= 5;
arr[0][2]= 8;
arr[0][3]= 9;
arr[0][4]= 4;
arr[1][0]= 2;
arr[1][1]= 1;
arr[1][2]= 0;
arr[1][3]= 2;
arr[1][4]= 4;
arr[2][0]= 6;
arr[2][1]= 7;
arr[2][2]= 4;
arr[2][3]= 5;
arr[2][4]= 5;
arr[3][0]= 4;
arr[3][1]= 8;
arr[3][2]= 9;
arr[3][3]= 1;
arr[3][4]= 2;
arr[4][0]= 4;
arr[4][1]= 5;
arr[4][2]= 2;
arr[4][3]= 1;
arr[4][4]= 9;
//convert 2 D array in 1 D array
for(i=0;i<5;i++){
printf("\n");
for(j=0;j<5;j++){
printf(" %d",arr[i][j]);
result[k++]=arr[i][j];
}
}
// sort 1 D array
for(i=0;i<25;i++){
for(j=0;j<24;j++){
if(result[j] > result[j+1]){
temp=result[j];
result[j]=result[j+1];
result[j+1]=temp;
}
}
}
/*
for(i=0;i<25;i++){
printf("\n%d",result[i]);
}*/
// convert 1 D array to 2 D array
i=0;
l=0;k=0;
while(i<25){
for(j=0;j<5;j++){
arr[k][j]=result[l];
l++;
}
k++;
i=i+5;
}
//Print matrix i.e 2D array
for(i=0;i<5;i++){
printf("\n");
for(j=0;j<5;j++){
printf(" %d",arr[i][j]);
}
}
}
This works !
#define COL 5
#define ROW 2
int main(){
int temp ;
int arr[2][5]= {2,15,26,14,12,18,1,2,3,4 };
int arr2[10] = {0};
int index = 0 ;
for(int t = 0 ; t<50 ; t++ ){
for (int i =0 ; i < ROW ; i++){
for( int j = 0; j < 5-1 ; j++){
if (arr[i][j] > arr[i][j+1]){
temp = arr[i][j];
arr[i][j] = arr[i][j+1];
arr[i][j+1] = temp;
}
}
//checking for
for( int k = 0 ; k < ROW-1 ; k++){
if (arr[k][COL-1] > arr[k+1][0]){
temp = arr[k][COL-1];
arr[k][COL-1] = arr[k+1][0];
arr[k+1][0] = temp ;
}
}
//---------
}
}
return 0 ;
}