So I am making an array of 52 ints, I add in random numbers and make sure there all different. If there is a duplicate I would generate a new number so that current index that am at is different. Where am I going wrong?
for(i=0; i < SIZE; i++){
deck[i] = (rand() % 52);
for(c= 0; c <= i; c++ ){
if(deck[c] == deck[i+1]){
deck[i] = (rand() % 52);
}
}
}
You can't try merely once for the random number in your inner for loop. After you generate it, you have to check all the previously generated items again to make sure you didn't generate the same random number as another item. This would do the trick (assuming SIZE = 52):
for(i=0; i < SIZE; i++){
deck[i] = (rand() % SIZE);
for(c= 0; c < i; c++ ){
if(deck[c] == deck[i]){
// Try another number
i--;
break;
}
}
}
That said, it's not a very fast solution (it could potentially never end). It's better to create an array of numbers 0 to 51 and then shuffle them by swapping two elements at a time. I'm not going to write that for you, but it's a standard way of doing it.
Totally the wrong approach. You don't want random numbers; you want precisely the numbers 0 to 51, in random order. To do this, fill the array with the values 0..51, and then shuffle it properly:
for (int i = 0; i < 52; i += 1) deck[i] = i;
for (int i = 52; i > 1; i -= 1) {
int j = rand() % i;
int temp = deck[j];
deck[j] = deck[i-1];
deck[i-1] = temp;
}
Related
#include<stdio.h>
#include<stdlib.h>
main()
{
int i,j,l,m,n;
j=0;
printf("\nenter 5 element single dimension array\n");
printf("enter shift rate\n");
scanf("%d",&n);
/* Here we take input from user that by what times user wants to rotate the array in left. */
int arr[5],arrb[n];
for(i=0;i<=4;i++){
scanf("%d",&arr[i]);
}
/* Here we have taken another array. */
for(i=0;i<=4;i++){
printf("%d",arr[i]);
}
for(i=0;i<n;i++){
arrb[j]=arr[i];
j++;
// These loop will shift array element to left by position which's entered by user.
}
printf("\n");
for(i=0;i<=3;i++){
arr[i]=arr[i+n];
}
for(i=0;i<=4;i++){
if(n==1 && i==4)
break;
if(n==2 && i==3)
break;
if(n==3 && i==2)
break;
printf("%d",arr[i]);
}
//To combine these two arrays. Make it look like single array instead of two
for(i=0;i<n;i++){
printf("%d",arrb[i]);
}
// Final sorted array will get printed here
}
Is it the efficeint program to rotate array in left direction?
Actually, very complicated, and some problems contained:
for(i = 0; i < n; i++)
{
arrb[j] = arr[i];
j++;
}
Why not simply:
for(i = 0; i < n; i++)
{
arrb[i] = arr[i];
}
There is no need for a second variable. Still, if n is greater than five, you get into trouble, as you will access arr out of its bounts (undefined behaviour!). At least, you should check the user input!
for(i = 0; i <=3 ; i++)
{
arr[i] = arr[i + n];
}
Same problem: last accessible index is 4 (four), so n must not exceed 1, or you again access the array out of bounds...
Those many 'if's within the printing loop for the first array cannot be efficient...
You can have it much, much simpler:
int arr[5], arrb[5];
// ^
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(i + n) % 5];
This does not cover negative values of n, though.
arrb[i] = arr[(((i + n) % 5) + 5) % 5];
would be safe even for negative values... All you need now for the output is:
for(int i = 0; i < 5; ++i)
printf("%d ", arrb[i]);
There would be one last point uncovered, though: if user enters for n a value greater than INT_MAX - 4, you get a signed integer overflow, which again is undefined behaviour!
We can again cover this by changing the index formula:
arrb[i] = arr[(5 + i + (n % 5)) % 5];
n % 5 is invariant, so we can move it out of the loop:
n %= 5;
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(5 + i + n) % 5];
Finally, if we make n positive already outside, we can spare the addition in the for loop.
n = ((n % 5) + 5) % 5;
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(i + n) % 5]; // my original formula again...
Last step is especially worth considering for very long running loops.
I think you want to do something like this (you should check that 0 <= n <= 5, too):
int b[5];
int k = 0;
for(i=0; i<5; i++){
if (i < 5 - n)
b[i] = arr[i+n];
else
{
b[i] = arr[k];
k++;
}
}
Array b is used to save the rotated matrix.
I do a code that will display to the screen 10 random numbers with no repetitions. I want to know if we can optimize the code or if you have a better and simple way in order to do this request.
Thanks !!
int main(){
int nbr = 0; srand(time(NULL));
int arr[10], i, j, flag;
for (i = 0; i < 10; i++)
{
do
{
nbr = rand() % 10 + 1;
flag = 1;
for (j = 0; j < i; j ++)
{
if (nbr == arr[j])
{
flag = 0;
break;
}
}
} while (!flag);
arr[i] = nbr;
}
for (i = 0; i < 10; i++)
{
printf("%5d", arr[i]);
}
system("PAUSE");
return 0;
}
So if i get what you're trying to do here it is:
generate an array of numbers between 1 and 10, in a random order (given rand() % 10 + 1)
instead of trial and error I'd suggest the following algorithm:
fill arr with 1...10
shuffle arr
this will run a lot faster
While I agree with the solution provided by Work of Artiz, this will result in the hard question to answer of when to stop shuffling.
To solve this you can use the following solution (which will use more memory, but less clock time):
1 Create an array temp having values 1..10 (ordered, not random)
2 Keep track of the length length of the array (10)
3 Generate a random index rand_i between 0 and length - 1
4 Copy temp[rand_i] to next position in your final array
5 Overwrite temp[rand_i] by temp[length-1]
6 Decrement length
7 Iterate Steps 3 - 6 until your array is filled (10 times)
This will both eliminate your excessive looping, and the problem of when to stop shuffling your array
EDIT: including code
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
int main(){
int nbr = 0; srand(time(NULL));
int arr[10], i, j, flag;
int temp[10] = {1,2,3,4,5,6,7,8,9,10};
int length = 10;
for (i = 0; i < 10; i++)
{
nbr = rand() % length; // Generate random index between 0 and length - 1
arr[i] = temp[nbr]; // Copy value from random index in temp to next index in arr
// Now since temp[nbr] is already taken (copied to arr) we should replace it by last available value in temp (temp[lenght])
// and in the next iteration we will create a value between 0 and length -1. This will keep values not yet taken from temp in
// contiguous order
temp[nbr] = temp[length-1];
length--;
}
for (i = 0; i < 10; i++)
{
printf("%5d", arr[i]);
}
return 0;
}
int main() //8th task
{
int longNum, shortNum, tempNum[5], i;
printf("Please enter 2 numbers (5 digits and 1 digit, ex: 12345 and 5)\n");
scanf("%d%d", &longNum, &shortNum);
for (i = 4; i >= 0; i--)
{
if (longNum % 10 != shortNum)
{
tempNum[i] = longNum % 10;
longNum /= 10;
}
else tempNum[i] = ; // Delete the digit that == shortNum.
}
for (i = 0; i < 5; i++)
{
printf("%d", tempNum[i]);
}
printf("\n");
return 0;
}
This program check if longNum has shortNum in it and suppose to remove the number (and his array slot) from longNum.
I've tried couple of things to make it work with no success.
I'd like to know what is the best way to do it (im not sure what the 'else' should be).
It is possible to skip all shortNum digits in the parsing loop. One more variable is needed to track number of deleted digits:
int n = 5;
for (i = 4; i >= 0; i--)
{
int tmp = longNum % 10;
longNum /= 10;
if (tmp != shortNum)
tempNum[--n] = tmp;
}
// here n is number of deleted digits
for (i = n; i < 5; i++)
{
printf("%d", tempNum[i]);
}
So, actually elements are not deleted from array. They are not written to that array. It is also possible to reverse elements order, so the first array element will be meaningful. Now if some element is skipped the first element of tempNum contains junk.
you need to skip the value that you don't want, and not insert it at all to the array.
int len = 0;
for (i = 4; i >= 0; i--)
{
if (longNum % 10 != shortNum)
{
tempNum[len] = longNum % 10;
len++;
}
longNum /= 10;
}
for (i = 0; i < len; i++)
{
printf("%d", tempNum[i]);
}
else tempNum[i] = ; this part is very wrong. You have to assing something like else tempNum[i] = 0;. And you can't actually delete anything from these arrays - they are not dynamic. I suggest you read up on dynamic arrays.
This question already has answers here:
Generating list of random numbers without duplication using C?
(5 answers)
Closed 8 years ago.
I want to generate random numbers between 1 and 13 without repetition
I used this method, but it doesn't make sure there is no reputation.
for ( i = 0; i < 13; i++)
{
array[i] = 1 + (rand() % 13);
}
Please help me.
C language
As a comment said, Fill an array with numbers 1 through 13 then shuffle the array.
int array[13];
for (int i = 0; i < 13; i++) { // fill array
array[i] = i;
printf("%d,", array[i]);
}
printf("\n done with population \n");
printf("here is the final array\n");
for (int i = 0; i < 13; i++) { // shuffle array
int temp = array[i];
int randomIndex = rand() % 13;
array[i] = array[randomIndex];
array[randomIndex] = temp;
}
for (int i = 0; i < 13; i++) { // print array
printf("%d,",array[i]);
}
Here is the sample output.
0,1,2,3,4,5,6,7,8,9,10,11,12,
done with population
here is the final array
11,4,5,6,10,8,7,1,0,9,2,12,3,
Note: I used the most basic sort I could. Use a better one if you want.
In the program that I'm writing, I currently have a for-loop that goes through an array, num[5], and checks to see if there are any 1's in that array, which looks like:
int counter = 0;
for (int i = 1; i <= 5; i++)
if (num[i] == 1)
counter++;
This works successfully, but I'm now trying to go through the array and see what the indices of the 1's in the program are. So, if I have 01001, I want to make an array that holds the positions of the 1's. The following is what I've tried so far:
int b[counter];
for (int k = 0; k <= counter; k++) {
for (i = 0; i <= 5; i++) {
if (num[i] == 1) {
b[k] = i;
}
}
}
but this doesn't produce the desired result. When I type the string in, say 1001, I get 444. Can someone tell me what I'm doing incorrectly?
For each value of k, for each occurrence of a 1, you're setting b[k] to the index of the 1. Thus each b[k] will have the index of the last 1.
Try:
int b[counter];
int k = 0;
for (i = 0; i <= 5; i++) {
if (num[i] == 1) {
b[k++] = i;
}
}
So, whenever it gets a 1, it assigns b[k] to the index and then increases k.
Then you should also use k, not counter, when trying to print out b.
The problem lies in this part of your code.
int b[counter];
for (int k = 0; k <= counter; k++) {
**for (i = 0; i <= 5; i++) {
if (num[i] == 1) {
b[k] = i;
}**
}
}
Suppose you get a 1 at the index 1 of the array as in 01001. You assign b[k] = 1;
This is perfectly valid. But as the loop continues you get another 1 at the index 4. Thus the command b[k] = 4; is again executed.
Note that your value of k is constant in both the statements and hence you get the array b as 44.
So what you need to do is break the inner for loop as soon as you get a 1.
Here is the modified code. You also need to keep a track of the iterator i and I have done that here using the variable- new_pos // new position.
int b[counter];
int new_pos=0; //to keep track of the iterator
for (int k = 0; k <= counter; k++) {
for (i = new_pos; i <= 5; i++) {
if (num[i] == 1) {
b[k] = i;
new_pos = i+1;
break;
}
}
}
The code provided by Dukeling is also perfect, but I am just giving another way to make your own code work.