I am trying to construct a 2D array for an assignment. I've used a nested for loop to construct the 2D array using scanf():
int width;
int height;
scanf("%d %d",&width,&height);
int array[width][height];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
However when I print the array, I can see that it has been constructed in a strange way, where all the numbers of the first line past a certain point are the first few numbers from the second line (instead of what they should be). The next lines after work fine.
Example:
Input:
6 2
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
The created array looks like this:
1 3 2 4 6 8 (<-- these last 4 numbers are the first 4 numbers of the second line)
2 4 6 8 0 2 (correct)
3 4 2 0 1 3 (correct)
Any ideas? Thanks a lot.
Your declaration of array
int array[width][height];
is wrong. The outer loop goes from 0 to height - 1, but array[i] can only go
from 0 to width - 1. The same applies for the inner loop. You swapped width
and height in the declaration of the array, it should be
int array[height][width];
Also note that for the matrix
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
the width is 6 and the height is 3, so the correct input should be
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
I compiled and run this code:
#include <stdio.h>
int main(void)
{
int width;
int height;
scanf("%d %d",&width,&height);
int array[height][width];
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
scanf("%d",&array[i][j]);
}
}
printf("----------------\n");
for (int i=0;i<height;i++){
for (int j=0;j<width;j++){
printf("%d ", array[i][j]);
}
printf("\n");
}
}
And the output is:
$ ./b
6 3
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
----------------
1 3 5 7 9 1
2 4 6 8 0 2
3 4 2 0 1 3
as you can see, now it's reading correctly. See https://ideone.com/OJjj0Y
Related
I have to create a Sudoku grid with the following pattern in C:
1 2 3 4
3 4 1 2
2 3 4 1
4 1 2 3
The first number in the top left corner (here: 1) must be an editable variable for a start value. There is another variable to create the grid with by the square size, in this example the square size is 2 and the 1 2 3 4 are in one square. 3 4 1 2 are in another square and so on...
If the start value is e.g. 3, the grid looks like this:
3 4 1 2
1 2 3 4
4 1 2 3
2 3 4 1
I noticed that there is a pattern: If the row number is odd, the new start value of the next row is the second last one. If the row number is even, the new start value of the next row is the last one. I tried to do it in C, but the even rows are cloning themselves. Note that arrays and pointers are not allowed here, only loops and other control-structures.
I tried the following approach, but the even rows are cloning themselves:
#include <stdio.h>
const int intSquareSize = 2;
const int intFieldLength = intSquareSize * intSquareSize;
int intStartValue = 3;
int main() {
int a = 0;
int b = 0;
int m = 0;
for (int intRowCounter = 1; intRowCounter <= intFieldLength; intRowCounter++) {
m = intFieldLength - 1;
for (int intColumnCounter = 1; intColumnCounter <= intFieldLength; intColumnCounter++) {
a = intStartValue + (intColumnCounter - 1);
b = a;
if (a > intFieldLength) {
a = intFieldLength - m;
m--;
}
if (intRowCounter % 2 == 0 && intColumnCounter == intFieldLength) {
intStartValue = a;
} else if (intRowCounter % 2 == 1 && intColumnCounter == (intFieldLength - 1)) {
intStartValue = b;
}
printf("%d\t", a);
}
printf("\n");
}
return 0;
}
What did I wrong and how can I fix it?
If the row number is odd, the new start value of the next row is the second last one. If the row number is even, the new start value of the next row is the last one.
I don't think it is helpful to think in terms of odd and even. The involved numbers are just symbols, and one could even replace them with distinct colors (for example). Odd/even is not a significant thing here, and it would certainly not play the same role in other board sizes.
The pattern I see is this:
In the first intSquareSize rows, the values shift horizontally (compared to the previous row) by intSquareSize. For example with intSquareSize=3 the first three rows could be:
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8
Notice the shift of 3 positions to the left at each next row.
Then the pattern for the next chunks of intSquareSize rows would be the same, but with one shift. So the complete 9x9 sudoku would look like this:
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
1 2 3 4 5 6 7 8 9
5 6 7 8 9 1 2 3 4
8 9 1 2 3 4 5 6 7
2 3 4 5 6 7 8 9 1
This is just one possible pattern you could follow. There could be others. But the following code will apply the above logic. Note I used your variables, but I prefer to use 0-based logic, so loop variables start at 0, and the value of a is also 0-based. Only at the time of printing 1 is added to that value, so it becomes 1-based:
int a = intStartValue - 1; // Move from 1-based to 0-based
for (int intBlockCounter = 0; intBlockCounter < intSquareSize; intBlockCounter++) {
for (int intRowCounter = 0; intRowCounter < intSquareSize; intRowCounter++) {
for (int intColumnCounter = 0; intColumnCounter < intFieldLength; intColumnCounter++) {
printf("%d\t", (a + 1)); // back to 1-based
a = (a + 1) % intFieldLength;
}
printf("\n");
a = (a + intSquareSize) % intFieldLength; // Shift within a block
}
a = (a + 1) % intFieldLength; // Shift between blocks
}
I have a file, with format:
Course - Grade Count - Grades
Programming 10 3 4 5 4 3 2 4 5 2 3
Mathematics 8 3 3 4 5 3 2 2 3
Physics 6 3 4 5 3 4 5
Design 6 5 4 5 3 2 4
Logistics 8 3 4 5 3 1 1 2 4
Ex: Course - Programming, Grade Count - 10 and Grades - 3 4 5 4 3 2 4 5 2 3
I already have
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define SIZE 70
int main(void)
{
char subject[SIZE];
int gradeCount;
int grades[SIZE];
FILE *fp = fopen("C:\\Project\\project.txt", "r"); //opening already created file
if (fp == NULL) {
perror("Error opening file");
return(-1);
}
for (int i = 0; i < SIZE; i++) {
fscanf(fp, "%s %d", &subject[i], &gradeCount);
printf("%s \n", &subject[i]);
//printf("%d \n", gradeCount);
for (int k = 0; k < gradeCount; k++)
{
fscanf(fp, "%d", &grades[k]);
// printf("%d \n" , grades[k]);
}
if (i == SIZE) {
break;
}
}
fclose(fp);
return 0;
}
I need to print out "Course", "Grade Count" and "Grades" without any problems, later on I need to make a search and so I need to separate them from each other, but that is not the case, now I will show you the outputs for all cases, when I output first "Subject/Course" then "Grade Count" and finally "Grades".
For Courses:
Programming
Mathematics
Physics
Design
Logistics
ogistics
gistics
istics
stics
tics
ics
cs
s
#
##
#
#
For Grade Count:
10
8
6
6
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
And for Grades:
3
4
5
4
3
2
4
5
2
3
3
3
4
5
3
2
2
3
3
4
5
3
4
5
5
4
5
3
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
3
4
5
3
1
1
2
4
In all cases, additional things are added to original stuff that should be printed out, I don't know where it comes from, I thought about pointers, but don't know much about them. Any suggestions?
Just need to print everything normally to normally search for everything (Courses, grade count and grades) later on.
You need to exit the loop early if it fails to read anything in. You can do that by checking the return value of fscanf. If the first call doesn't return 2, you know that it didn't read in 2 values and can break out of the loop.
You're also calling fscanf and printf incorrectly for dealing with a string. You are moving the starting point of where you read into/print from, which isn't needed and reduces the maximum space available to you.
Updated code looks something like this
for (int i = 0; i < SIZE; i++) {
if(fscanf(fp, "%s %d", subject, &gradeCount) != 2) {
break;
}
printf("%s ", subject);
//printf("%d \n", gradeCount);
for (int k = 0; k < gradeCount; k++)
{
fscanf(fp, "%d", &grades[k]);
// printf("%d" , grades[k]);
}
}
So I have to take an ID array
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
And perform weighted quick union on it. I have to perform the operations 9-0, 3-4, 5-8, 7-2, 2-1, 5-7, 0-3, and 4-2. Here's what I did to the array for these operations:
9-0
0 1 2 3 4 5 6 7 8 9
9 1 2 3 4 5 6 7 8 9
3-4
0 1 2 3 4 5 6 7 8 9
9 1 2 3 3 5 6 7 8 9
5-8
0 1 2 3 4 5 6 7 8 9
9 1 2 3 3 5 6 7 5 9
7-2
0 1 2 3 4 5 6 7 8 9
9 1 7 3 3 5 6 7 5 9
2-1
0 1 2 3 4 5 6 7 8 9
9 7 7 3 3 5 6 7 5 9
5-7
0 1 2 3 4 5 6 7 8 9
9 7 7 3 3 7 6 7 5 9
0-3
0 1 2 3 4 5 6 7 8 9
9 7 7 3 3 7 6 7 5 3
4-2
0 1 2 3 4 5 6 7 8 9
9 7 7 3 3 7 6 3 5 9
The problem is that the ID array operations are different depending on if you're using quick find or quick union or weighted quick union. So would this be right for weighted quick union? Here's the code I'm using for weighted quick union:
public class WeightedQuickUnionUF
{
private int[] id; // parent link (site indexed)
private int[] sz; // size of component for roots (site indexed)
private int count; // number of components
public WeightedQuickUnionUF(int N)
{
count = N;
id = new int[N];
for (int i = 0; i < N; i++) id[i] = i;
sz = new int[N];
for (int i = 0; i < N; i++) sz[i] = 1;
}
public int count()
{ return count; }
public boolean connected(int p, int q)
{ return find(p) == find(q); }
private int find(int p)
{ // Follow links to find a root.
while (p != id[p]) p = id[p];
return p;
}
public void union(int p, int q)
{
int i = find(p);
int j = find(q);
if (i == j) return;
// Make smaller root point to larger one.
if (sz[i] < sz[j]) { id[i] = j; sz[j] += sz[i]; }
else { id[j] = i; sz[i] += sz[j]; }
count--;
}
}
The code for weighted quick union shows you how it works but basically it's a type of Union-find where you connect two trees together. With weighted quick union you always connected the smaller tree to the larger one. The ID array is a representation of a tree where there's numbers on the top row and bottom row. If the number on top matches the bottom then that number is a root in the forest but for example if the top number is a 9 and the bottom number is 0 then it means 9 is a child of 0. The ID array starts out with 9 single node trees and operations like 9-0 connects two trees together.
I have a square 2d array of values, where each row is identical, and where each element of row is one bigger than the last. For example:
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
I want to filter them, such that I can make a diamond as such:
1
1 2 3
1 2 3 4 5
1 2 3 4 5 6 7
1 2 3 4 5
1 2 3
1
Notice how the first part of the array is used, no matter how many elements are to be printed on that line. Also, spacing doesn't matter. I spaced them to show the diamond.
I know how to filter the top right "chunk" out, using j-i<(j/2). This will convert the original square into:
1
1 2 3
1 2 3 4 5
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
How can I get the bottom right "chunk" to filter out also? What additional condition can I impose on the values?
Presuming you have found out and stored the length of the "side" of the square already then you could use something like below. However, if your square has an even length then it will not work (can't produce a diamond in this way from an even side length square).
The following is pseudo-code so you will need to adapt it for your language. I've also used 0-indexed arrays and presumed square is a 2D array.
for (i=0, i<length, i++)
{
for (j=0, j<Length, j++)
{
if (i < length/2)
{
if (j < length/2 AND j <= i)
print square[i][j]
}
}
else
{
if (j < length/2 AND j <= (length - i))
{
print square[i][j]
}
}
}
print newline
}
Begining with an ordered array
[1, 2, 3, 4, 5, 6, 8, 9, 10]
How would be the way to get every iteration the following results?
1 2 3 4 5 6 7 8 9 10
1 3 4 5 6 7 8 9 10 2
1 4 5 6 7 8 9 10 2 3
1 5 6 7 8 9 10 2 3 4
1 6 7 8 9 10 2 3 4 5
1 7 8 9 10 2 3 4 5 6
1 8 9 10 2 3 4 5 6 7
1 9 10 2 3 4 5 6 7 8
1 10 2 3 4 5 6 7 8 9
#include <stdio.h>
#define MAX 10
int a[MAX], i,j,cnt=2;
main (){
for (i=0; i<MAX; i++){
a[i]= i+1;
}
for (i=0; i<MAX; i++) {
printf ("%d ", a[i]);
}
printf ("\n");
for (j=0; j < MAX-2;j++){
a[0]=1;
for (i=1; i < MAX-1; i++){
if (a[i]%MAX != 0){
a[i]= a[i] + 1;
}else{
if (a[i]==10) {
//printf ("a[%d]: %d \t ** %d\n", i , a[i] ,cnt);
//a[i-1]= i;
a[i] = cnt;
}
}
}
for (i=0; i<MAX; i++) {
printf ("%d ", a[i]);
}
printf ("\n");
}
}
Now I almost get it but the last column is not right, What should I do?
1 2 3 4 5 6 7 8 9 10
1 3 4 5 6 7 8 9 10 10
1 4 5 6 7 8 9 10 2 10
1 5 6 7 8 9 10 2 3 10
1 6 7 8 9 10 2 3 4 10
1 7 8 9 10 2 3 4 5 10
1 8 9 10 2 3 4 5 6 10
1 9 10 2 3 4 5 6 7 10
1 10 2 3 4 5 6 7 8 10
C arrays are indexed from 0. So when you access elements from 1 to MAX, you are running off the end of the array.
Have your loops go from 0 to MAX-1. Customary way to write it is
for (i=0 ; i < MAX ; ++i)
...so anybody reading your code can immediately prove that the array index never equals MAX.
Well, at a minimum, arrays in C are zero based so you are writing past the end of the array. For an array declared int foo[MAX] valid elements are from foo[0]…foo[MAX-1]
Specifically a[MAX] might well reference the memory location that the variable i uses, causing the loop to reset when it attempt to overwrite a[MAX].
Either shift everything down by one, or declare your array MAX+1 and ignore the zero bit.
Oh, and you should not need to set a[1]=1; every time.