I am trying to get my program to count down after counting up to ten. I have tried to alter the code from counting up to make it count down to no avail.
#include <stdio.h>
void count(int k)
{
if (k > 0) {
count(-k + 1);
printf("%d", k);
}
else {
if (k == 0)
{
printf("%d,", k);
}
else {
count(k + 1);
printf("%d,", -k);
}
}
}
int main(int argc, char ** argv)
{
count(10);
getchar();
return 0;
}
Here is a simple example of the recursion which does this, illustrating Eugene's comment:
#include <stdio.h>
void count(int n) {
if (n > 10) {
printf("\n");
return;
}
printf("%d ", n);
count(n+1);
printf("%d ", n);
}
int main() {
count(0);
printf("\n");
return 0;
}
it counts up on the way into recursion and counts down while it exits it. Actually on the way down it only re-prints the state which it was before diving into the next level:
0 1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1 0
The function can be easy implemented if to use a static local variable inside it. For example.
#include <stdio.h>
void count(unsigned int n)
{
static unsigned int m;
printf("%u ", m);
if (n != m)
{
++m;
count(n);
--m;
printf("%u ", m);
}
}
int main( void )
{
const unsigned int N = 10;
unsigned int i = 0;
do
{
count(i);
putchar('\n');
} while (i++ != N);
return 0;
}
The program output is
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 5 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 1 0
0 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1 0
Within the function the static variable m behaves as an index in a for loop (or there will be more suitable a do-while loop).
At first it is initialized implicitly by zero (as any static variable)
static unsigned int m;
You can use the initializer explicitly if you want
static unsigned int m = 0;
then it is changed from 0 to n and afterward backward from n again to 0.
++m; // changing from 0 to n
count(n);
--m; // changing from n to 0
Related
Print all n! permutations of the number 1,2,3,...,n.
Example: Input: 3
Output: 1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
Following is my approach. My program is not working for inputs greater than 3. I understand the logic why it is not working , but I am unable to translate that logic into a code block to overcome that issue.
#include <stdio.h>
int permute(int n)
{
int a[n];
int i,j,k,store;
for(i=0;i<n;i++)
a[i]=i+1;
for(i=1;i<=n;i++)
{
for(j=0;j<n-1;j++)
{
store=a[j+1];
a[j+1]=a[j];
a[j]=store;
for(k=0;k<n;k++)
printf("%d ",a[k]);
printf("\n");
}
}
}
int main()
{
int n;
scanf("%d",&n);
permute(n);
return 0;
}
Following is the output for n as 4:
We can clearly see that some permutation are missing, and I know exactly the fault in my code. But I am unable to fix it.( I am a beginner , hence I don't know much advanced C libraries or functions)
One solution consists in calling the function recursively: you set the first number (n possible choices), then call the function for a size n-1.
Output, for n=4
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 3 2
1 4 2 3
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 3 1
2 4 1 3
3 2 1 4
3 2 4 1
3 1 2 4
3 1 4 2
3 4 1 2
3 4 2 1
4 2 3 1
4 2 1 3
4 3 2 1
4 3 1 2
4 1 3 2
4 1 2 3
#include <stdio.h>
#include <stdlib.h>
void swap (int *i, int *j) {
int temp = *i;
*i = *j;
*j = temp;
}
void permute(int index, int* arr, int n) {
if (index == n-1) {
for (int k = 0; k < n; ++k) {
printf ("%d ", arr[k]);
}
printf ("\n");
return;
}
for (int i = index; i < n; i++) {
swap (arr + index, arr + i);
permute (index+1, arr, n);
swap (arr + i, arr + index);
}
return;
}
int main()
{
int n;
if (scanf("%d",&n) != 1) exit (1);
int arr[n];
for (int i = 0; i < n; ++i) arr[i] = i+1;
permute(0, arr, n);
return 0;
}
Given a number — 4 — I have to 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 main code I wrote has an error for direction. According to me, I wrote the correct code, but it only works for 1 and 3 and loops between it — 1 3 1 3 1 3.... it is supposed to work 0 1 2 3 0 1 2 3 0 1.... accordingly in a spiral manner. I would really appreciate if you could help me!
The code I wrote is as follows:
int main()
{
int n;
scanf("%d", &n);
int x=n+(n-1);
int size=x*x;
int t,b,l,r;
t=0;l=0;
b=x-1;
r=x-1;
int i,j;
int res[size];
int index=0;
int dir=0;
while(t<=b && l<=r){
if(dir==0){
for(i=l;i<r;i++){
res[index]=n;
index++;
}
t++;
dir=1;
}
else if (dir==1) {
for(i=t;i<b;i++){
res[index]=n;
index++;
}
r--;
dir=2;
}
else if (dir==2) {
for(i=r;i>l;i--){
res[index]=n;
index++;
}
b--;
dir=3;
}
else if(dir==3){
for(i=b;i>l;i++){
res[index]=n;
index++;
}
l++;
dir=0;
}
n--;
dir=(dir+1)%4;
// printf("%d",dir); if u want to check
}
/* yet to proceed
for(i=0;i<x;i++){
for(j=i;j<x;j=j+x){
printf("%d ",res[j]);
}
printf("\n");
}
*/
return 0;
}
I tried editing the loop for counter so that the last line of printf is non negative but I'm not too sure which part of the loop to edit.
#include <stdio.h>
void printPatternHere(int height);
int main() {
int height;
printf("Enter the number of rows: ");
scanf("%d", &height);
printf("printPattern: \n");
printPatternHere(height);
return 0;
}
void printPatternHere(int height) {
int n, c, row, t = 1;
int counter = 0;
for (row = 1; row <= height; row++) {
for (c = 1; c <= height - row; c++)
t = row; // start each line with row number
for (c = 1; c <= row; c++) {
counter++;
if (counter > 9) {
t = t % 10;
}
printf("%d ", t);
t++;
}
for (c = 1 ; c < row; c++) {
t--; //reset row number
}
printf("\n");
}
}
If userinput is 14, its output should be something like this:
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
6 7 8 9 0 1
7 8 9 0 1 2 3
8 9 0 1 2 3 4 5
9 0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0 1
2 3 4 5 6 7 8 9 0 1 2 3
3 4 5 6 7 8 9 0 1 2 3 4 5
4 5 6 7 8 9 0 1 2 3 4 5 6 7
Your function is cluttered too much, for loops are more than enough. Code below will be enough. Better to debug to grasp the idea.
// in "main"
printPattern3(height); // wrong function name!
printPatternHere(height); // corrected.
void printPatternHere(int height) {
int c, row;
for (row = 1; row <= height; row++) {
for (c = 0; c < row; c++) {
printf("%d ", (c + row) % 10);
}
printf("\n");
}
}
Your code is too complicated. The output of the pyramid can be done simpler.
For example this loop
for (c = 1; c <= height - row; c++)
t = row; // start each line with row number
does not make a sense.
Here you are.
#include <stdio.h>
int main(void)
{
while ( 1 )
{
const unsigned int Base = 10;
printf( "Enter a non-negative number (0 - exit): " );
unsigned int n;
if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
putchar( '\n' );
for ( unsigned int i = 0; i < n; i++ )
{
for ( unsigned int j = 0; j < i + 1; j++ )
{
printf( "%u ", ( j + i + 1 ) % Base );
}
putchar( '\n' );
}
putchar( '\n' );
}
return 0;
}
The program output might look like
Enter a non-negative number (0 - exit): 14
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
6 7 8 9 0 1
7 8 9 0 1 2 3
8 9 0 1 2 3 4 5
9 0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 0 1
2 3 4 5 6 7 8 9 0 1 2 3
3 4 5 6 7 8 9 0 1 2 3 4 5
4 5 6 7 8 9 0 1 2 3 4 5 6 7
Enter a non-negative number (0 - exit): 0
If you want to write a separate function that will output the pattern then the program can look the following way.
#include <stdio.h>
FILE * display_pyramid( unsigned int n, FILE *fp )
{
const unsigned int Base = 10;
for ( unsigned int i = 0; i < n; i++ )
{
for ( unsigned int j = 0; j < i + 1; j++ )
{
fprintf( fp, "%u ", ( j + i + 1 ) % Base );
}
fputc( '\n', fp );
}
return fp;
}
int main(void)
{
while ( 1 )
{
printf( "Enter a non-negative number (0 - exit): " );
unsigned int n;
if ( scanf( "%u", &n ) != 1 || n == 0 ) break;
putchar( '\n' );
fputc( '\n', display_pyramid( n, stdout ) );
}
return 0;
}
I need to print all possible series that their sum is equal to N;
for example is n == 4 the output should be:
[1, 1, 1, 1]
[1, 1, 2]
[1, 2, 1]
[1, 3]
[2, 1, 1]
[2, 2]
[3, 1]
[4]
My way of thinking to solve this problem was:
print the series that the number i is not in the series
print the series that the number i is in the series, now need to find the sum of N-i.
my code:
#include <stdio.h>
#include <stdlib.h>
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
{
printf(" %d ", arr[i]);
}
printf("\n");
}
void printAllHelper(int* a,int size, int sum, int used,int index) {
if (sum == 0) {
a -= used;
printArr(a, used);
}
else if (sum < 0 || index == size)
{
return;
}
else {
for(int i = 1 ; i <= size ; i ++)
{
printAllHelper(a, size, sum, used, index + 1);
if (i <= sum)
{
*a = i;
}
printAllHelper(a+1, size, sum -i, used +1, index + 1);
}
}
}
void printAll(int num) {
int* myArray = (int*)malloc(num * sizeof(int));
printAllHelper(myArray,num,num,0,0);
}
void main() {
printAll(4);
}
my output:
3 1
3 1
3 1
3 1
3 1
3 1
3 1
3 1
3 1
4
1 3
4
2 2
4
3 1
4
4
1 3
1 1 2
1 3
1 2 1
1 3
1 3
1 3
4
1 3
4
2 2
4
3 1
4
4
2 2
2 1 1
2 2
2 2
2 2
2 2
4
1 3
4
2 2
4
3 1
4
4
3 1
3 1
3 1
3 1
3 1
4
1 3
4
2 2
4
3 1
4
4
4
4
Please try to explain to me your way of thinking, and how you approach this kind of problem, I want to be the very best like no one ever was :(.....
Your reasoning is not quite correct, but your code is almost right. The loop in your else part should be
for(int i = 1 ; i <= sum ; i ++) {
*a = i;
printAllHelper(a+1, size, sum-i, used+1, index+1);
}
With this, I get the output
1 1 1 1
1 1 2
1 2 1
1 3
2 1 1
2 2
3 1
4
The idea is basically: "The numbers sum to sum if the first number i is any number from 1 to sum and the rest of the numbers sum to sum - i."
Also, note that your code shows some room for improvement, e.g. the used and index variables seem a bit redundant. And with not adding numbers larger than sum or smaller than 1, the check whether sum < 0 || index == size is not necessary, either. Thus you also do not need the size parameter. Your printAllHelper could be simplified to something like this:
void printAllHelper(int* a, int sum, int index) {
if (sum == 0) {
printArr(a, index);
} else {
for(int i = 1 ; i <= sum ; i++) {
a[index] = i;
printAllHelper(a, sum-i, index+1);
}
}
}
(Note: C is not my native language, if you see more things to improve, please comment.)
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 6 years ago.
Improve this question
I am having trouble with implementing the sort function on pset3. I have used the GDB and found that my sort function does not sort anything. I am not sure if there is a syntax issue, or if the logic is a bit screwed up.
void sort(int values[], int n)
{
for (int k = 0; k < n; k++)
{
for (int j = 0; j < n; j++)
{
if (values[k] >= values[j])
{
int temp = values[k];
values[k] = values[j];
values[j] = temp;
}
}
}
}
You are close, but your loops are not quite right - change:
for (int k = 0; k < n; k++)
{
for (int j = 0; j < n; j++)
{
to:
for (int k = 0; k < n - 1; k++)
{
for (int j = k + 1; j < n; j++)
{
To understand why you need to make this change, consider that the inner loop (j) need only compare elements above index k with the current element at index k. So the outer loop (k) needs to iterate from 0 to n - 2 (one less than the last element), and for each outer loop iteration the inner loop needs to iterate from k + 1 (first element above k) to n - 1 (the last element).
NOTE: by pure chance it seems that the original code does appear to work correctly, even though it appears at first glance that it shouldn't. I have tested it with various edge cases and even though it performs many redundant swaps, the final result always seems to be sorted (suprisingly though the output is in descending order whereas the fixed code generates results in ascending order, as expected). Credit to Jonathan Leffler for spotting this - see his answer and demo program.
One other minor point -- this test:
if (values[k] >= values[j])
should really just be:
if (values[k] > values[j])
It's not incorrect as it stands (the code will still work), but there is no point in swapping elements that are equal, so it's somewhat inefficient as written.
I took your code and converted into a complete program. It's larger than an MCVE because it has support code for shuffling arrays, and for printing results, as well as a main() that exercises these, of course.
#include <stdio.h>
#include <stdlib.h>
static int rand_int(int n)
{
int limit = RAND_MAX - RAND_MAX % n;
int rnd;
while ((rnd = rand()) >= limit)
;
return rnd % n;
}
static void shuffle(int *array, int n)
{
for (int i = n - 1; i > 0; i--)
{
int j = rand_int(i + 1);
int tmp = array[j];
array[j] = array[i];
array[i] = tmp;
}
}
static void print_array(int n, int a[n])
{
for (int i = 0; i < n; i++)
printf(" %d", a[i]);
putchar('\n');
}
static void sort(int values[], int n)
{
for (int k = 0; k < n; k++)
{
for (int j = 0; j < n; j++)
{
if (values[k] >= values[j])
{
int temp = values[k];
values[k] = values[j];
values[j] = temp;
}
}
}
}
int main(int argc, char **argv)
{
if (argc > 1)
{
long l = strtol(argv[1], 0, 0);
unsigned u = (unsigned)l;
printf("Seed: %u\n", u);
srand(u);
}
int data3[3] = { 3, 1, 2 };
print_array(3, data3);
sort(data3, 3);
print_array(3, data3);
int data5[5] = { 0, 2, 6, 1, 5, };
for (int i = 0; i < 5; i++)
{
shuffle(data5, 5);
print_array(5, data5);
sort(data5, 5);
print_array(5, data5);
}
int data9[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
for (int i = 0; i < 9; i++)
{
shuffle(data9, 9);
print_array(9, data9);
sort(data9, 9);
print_array(9, data9);
}
return 0;
}
The shuffle code implements a Fisher-Yates shuffle, and is
based on code from an answer by Roland Illig. If invoked without a seed argument, it generates the same output each time.
Code compiled and tested on macOS Sierra 10.12.1 with GCC 6.2.0.
An example output:
Seed: 123456789
3 1 2
3 2 1
6 0 1 5 2
6 5 2 1 0
0 6 1 2 5
6 5 2 1 0
0 1 2 6 5
6 5 2 1 0
5 0 6 1 2
6 5 2 1 0
1 6 5 2 0
6 5 2 1 0
0 4 8 3 7 5 1 6 2
8 7 6 5 4 3 2 1 0
7 4 0 5 6 8 3 2 1
8 7 6 5 4 3 2 1 0
1 2 7 5 0 8 3 6 4
8 7 6 5 4 3 2 1 0
3 8 7 5 2 1 0 6 4
8 7 6 5 4 3 2 1 0
1 4 2 6 3 0 7 5 8
8 7 6 5 4 3 2 1 0
2 3 7 4 8 0 5 6 1
8 7 6 5 4 3 2 1 0
3 4 5 8 6 2 0 7 1
8 7 6 5 4 3 2 1 0
3 6 7 4 8 2 5 1 0
8 7 6 5 4 3 2 1 0
0 8 7 3 4 6 5 1 2
8 7 6 5 4 3 2 1 0
This shows the data being sorted in descending order every time, despite different randomized inputs.