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;
}
Related
I have just started using C and I am currently working on calculating primes using the wikipedia algorithm here:
algorithm Sieve of Eratosthenes is
input: an integer n > 1.
output: all prime numbers from 2 through n.
let A be an array of Boolean values, indexed by integers 2 to n,
initially all set to true.
for i = 2, 3, 4, ..., not exceeding √n do
if A[i] is true
for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n do
A[j] := false
return all i such that A[i] is true.
When I try implementing what I think turns out like the code above, I get what I believe is an 'infinite loop', where might I have gone wrong?
#include <stdio.h>
#include <stdlib.h>
int main(void) {
//create empty array to store values
int isPrime[] = {};
//set a large number
int n = 1000;
//create for loop
for(int i = 2; i < n; i++){
///create another for loop taking the exponents
for(int j = i; j < pow(i, 2); j++){
//if i is equal to j is true then return those values true
if(isPrime[i] == isPrime[j]){
printf("%d", isPrime[i]);
}
}
}
There's many errors in your code including an empty array, use of pow (which is a floating-point function) and numerous logic errors that deviate from the wikipedia example code.
Here's a corrected, working version that follows the logic of the wikipedia version, with a loop afterwards to print out all the primes less than n:
#include <stdio.h>
int main(void) {
int n = 1000;
int isPrime[n];
for (int i = 0; i < n; i++) {
isPrime[i] = 1;
}
for (int i = 2; i * i < n; i++) {
if (isPrime[i]) {
for (int j = i * i; j < n; j += i) {
isPrime[j] = 0;
}
}
}
for (int i = 2; i < n; i++) {
if (isPrime[i]) {
printf("%d ", i);
}
}
printf("\n");
}
(Note that a small deviation from the wikipedia algorithm is that this prints primes less than n rather than primes less than or equal to n).
I am learning C and I keep mixing up array and matrix. and I can't seem to understand where I am doing wrong with my code. I have made 2 different version of it and the only feedback I am getting is, I have mixed array with matrix.
I was wondering if anyone could help me understand exactly where I went wrong, since I am not getting anywhere with my prof and his explanation about it. (Sorry in advanced)
This is the first code, But both of them are really similar.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int array[100];
int max = 0, min = 1;
int fro;
fro = (int)time(NULL);
srand(fro);
for (int i = 0; i < 100; i++) {
array[i] = rand() % (400 + 1 - 100) + 100;
if (array[i] <= min) {
min = array[i];
}
if (array[i] >= max)
max = array[i];
}
printf("Array\n");
for (int row = 0; row < 10; row++) {
for (int col = 0; col < 10; col++) {
printf(" %i ", array[row * 10 + col]);
}
printf("\n");
}
return 0;
}
This is another version, but I got similar feedback as the first one.
I am mixing up array and matrix..
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int tal[10][10];
int array[100];
int max = 0, min = 1;
int fro;
fro = (int)time(NULL);
srand(fro);
// Version 3 of the array slump generation
for (int row = 0; row < 10; row++) {
//printf("Yttre loop row %i\n", row);
for (int col = 0; col < 10; col++) { // row = 1, col = 0
tal[row][col] = rand() % (400 + 1 - 100) + 100;
// printf("tal[%i][%i] = %i\n", row, col, tal[row][col]);
// Get max
if (tal[row][col] >= max) {
max = tal[row][col];
}
// Get min
if (tal[row][col] <= min) {
min = tal[row][col];
}
// printf("tal[%i][%i] = %i\n", row, col, tal[row][col]);
}
}
// Printar ut hela matrisen i row-major order
for (int row = 0; row < 10; row++) {
for (int col = 0; col < 10; col++) {
printf(" %i ", tal[row][col]);
}
printf("\n");
}
return 0;
}
I mean both of them works and I think I am using array :/
We haven't even gone trough matrix yet...
The notion of a matrix is not an ISO C concept. However, you can use an array to denote a matrix by imposing imposing additional linear algebra constraints. There are several linear algebra libraries for C. However, be prepared for some people use the term matrix loosely to mean any two-dimensional array, because they look similar in most presentations.
Your first example is an int array[100] which stores array 100 of int. You then access it by 10 row and 10 col; since 10 * 10 = 100 <= 100 that you've reserved for it, this is entirely valid. One might prefer this representation because it is explicitly contiguous, but has no way for the compiler to bounds-check.
Your second example is int tal[10][10], which is array 10 of array 10 of int, all in the same block of memory. It's the same 100-element array, but accessed differently, as you've done. This is also valid, and I think more what your teacher was asking.
The one you treat more like a matrix will be more like a matrix.
I have puzzles that looks like this:
=== ====
=== ===
=== ====
left edge has length from 0 to 10 000 and right also, middle part is from 1 to 10 000.
So the question is if i can build a rectangle? like first puzzle has length of left edge equal to 0 and the last puzzle has right edge of length 0 and in the middle they fit perfectly?
I am given the number of puzzle i have and their params like this:
6
1 9 2
0 3 1
0 4 1
8 9 0
2 9 0
1 5 0
and result can be any of that:
2
0 3 1
1 5 0
or
3
0 3 1
1 9 2
2 9 0
or
2
0 4 1
1 5 0
But if there is no result i have to printf("no result")
I have to do this in C, I thought about doing some tree and searching it with BFS where vertices would have edge lengths and edge would have middle length and when reached 0 i would go all way up and collect numbers but it's hard to code. So i decided to do recursion but im also stuck:
#include<stdio.h>
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int tab[a][3];//array for puzzles
int result[][3];//result array
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
int findx(int x, int a, int *tab[], int *result[], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[a][0] == x){//there i look for puzzles with x length at start
if(tab[a][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[a][m];
}
return print_result(result, k);//we will return result go to print_result function
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[a][m];
k += 1;
}
findx(tab[a][2], a, tab, result, k);
}
}
}
printf("no result");
}
int print_result(int *result[], int k){
printf("%d", &k);//how many puzzles i have
printf("\n");
for(int i = 0; i < k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", &result[i][j]);
}
printf("\n");//...in separate lines
}
}
I have an error that result array can't look like this int result[][3] because of of [] but I don't know how many puzzles I'm gonna use so?... and I have implicit declaration for both of my functions. Guys please help, I dont know much about C and its super hard to solve this problem.
I'm not sure I understand the overall logic of the problem, but you definitely are in need of some variable sized containers for result AND tab. Arrays are fixed size and must be defined at compile time. The following should at least compile without warnings:
#include<stdio.h>
#include<stdlib.h>
void print_result(int (*result)[3], int k){
printf("%d", k);//how many puzzles i have
printf("\n");
for(int i = 0; i <= k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", result[i][j]);
}
printf("\n");//...in separate lines
}
}
void findx(int x, int a, int (*tab)[3], int (*result)[3], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[i][0] == x){//there i look for puzzles with x length at start
if(tab[i][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[i][m];
}
print_result(result, k);//we will return result go to print_result function
return;
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[i][m];
k += 1;
///** Increase size of result **/
//int (*newptr)[3] = realloc(result, (k+1) * sizeof(int[3]));
//if (newptr)
// result = newptr;
}
findx(tab[i][2], a, tab, result, k);
}
}
}
printf("no result\n");
}
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int (*tab)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int (*result)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
Note that I changed the tab and result types to (*int)[3]. Due to order of operations, we need parentheses here. Because they are variable size, they require dynamic memory allocations. In the interest of brevity and readability, I did not check the returned values of malloc or realloc. In practice, you should be checking that the returned pointer is not NULL. Because we are using dynamic memory allocations, we should also use free if you plan on doing anything else with this program. Otherwise, it doesn't really matter because exiting the program will free the resources anyway. You actually don't want to free. because we are passing a pointer by value to findx and the realloc can change the address, it may come back with a different address. Also, take note that I needed to include <stdlib.h> for the dynamic memory allocations.
Additional Issues
Your functions print_results and findx are not declared when you call them in main. Your function either need to be above main or have "function prototypes" above main.
In the printfs you do not need the &. You do not want to send the address of the variable to printf. You want to send what will actually be printed.
Now what?
The program still does not provide you with the correct results. It simply outputs 0 as the result every time. This should at least give you a starting point. By changing this line in print_results:
for(int i = 0; i < k; i++){//printing puzzles...
to
for(int i = 0; i <= k; i++){//printing puzzles...
I was at least able to output 0 0 0. This seems more correct because if k is 0, we don't loop at all.
#include<stdio.h>
void findx(int x, int a, int tab[a][3], int result[200000][3], int puzzlesinresult) { //i am looking for puzzle with length x on start
for (int i = 0; i < a; i++) {
if (tab[i][0] == x) { //there i look for puzzles with x length at start
if (tab[i][2] == 0) { //if i find such puzzle i check if this is puzzle with edge length zero at the end
for (int m = 0; m < 3; m++) { //this for loop add to my result array last puzzle
result[puzzlesinresult][m] = tab[i][m];
}
return print_result(result, puzzlesinresult); //we will return result go to print_result function
} else { //if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
while (result[puzzlesinresult - 1][2] != tab[i][0] && puzzlesinresult > 0) {
puzzlesinresult -= 1;
}
int isusedpuzzle = 0;
for (int j = 0; j < puzzlesinresult; j++) {
if (result[j][0] == tab[i][0] && result[j][1] == tab[i][1] && result[j][2] == tab[i][2]) {
isusedpuzzle = 1;
} else {
//pass
}
}
if (isusedpuzzle == 0) {
for (int m = 0; m < 3; m++) {
result[puzzlesinresult][m] = tab[i][m];
}
puzzlesinresult += 1;
findx(tab[i][2], a, tab, result, puzzlesinresult);
}
}
}
}
}
void print_result(int result[200000][3], int puzzlesinresult) {
printf("%d\n", puzzlesinresult + 1); //how many puzzles i have
for (int i = 0; i < puzzlesinresult + 1; i++) { //printing puzzles...
for (int j = 0; j < 3; j++) {
printf("%d ", result[i][j]);
}
printf("\n"); //...in separate lines
}
exit(0);
}
int main() {
int a;
scanf("%d", & a); //here i get how many puzzles i have
int tab[a][3]; //array for puzzles
int result[100][3]; //result array
int puzzlesinresult = 0; //this will help me track how many puzzles has my result array
for (int i = 0; i < a; i++) { //upload puzzles to array
for (int j = 0; j < 3; j++) {
scanf("%d", & tab[i][j]);
}
}
for (int i = 0; i < a; i++) { //here i delete puzzles that doesnt contribute anything like 1 x 1,2 x 2,..
if (tab[i][0] == tab[i][2] && tab[i][0] != 0) {
for (int p = i; p < a; p++) {
for (int j = 0; j < 3; j++) {
tab[p][j] = tab[p + 1][j];
}
}
}
}
findx(0, a, tab, result, puzzlesinresult); //start of recursion, because start has to be length 0
printf("NONE");
}
This returns sometimes correct result. If you can find when this program fails I would rly appreciate sharing those cases with me :)
#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.
Hi I have problem with initializing my function for printing an error message if some numbers in an array are same.
#include<stdio.h>
#include<stdlib.h>
void printRepeating(int arr[], int size)
{
int i, j;
for(i = 0; i < size; i++)
for(j = i+1; j < size; j++)
if(arr[i] == arr[j])
printf("Wrong input. Same numbers in array!\n");
}
int main()
{
int arr[200],i;
int res, num;
while((res = scanf("%d", &num)) == 1)
{
arr[i++] = num;
if(num == 0){
break;
}
}
for(i = 0; i < arr[i]; i++)
printf("%d ", arr[i]);
printf("\n");
int arr_size = sizeof(arr[i])/sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
If I scan 1 2 3 1 4 5 0, my function printRepeating wont start nevertheless I have numbers 1 1 that are same in the array, Why ? And another problem is when I type 1 2 3 1 5 0 it only prints 1 2 3 and for example I when I scan 1 2 3 4 5 6 7 8 9 0 it prints all numbers except for 0.
There are multiple issues in your code. First, initialize i to be 0, and declare a new variable j alongside i,
int arr[200], i = 0, j;
The size of your array would simply be i, which you increment every time you insert an element in the array, and change this,
for(i = 0; i < arr[i]; i++)
printf("%d ", arr[i]);
to this
for(j = 0; j < i; j++)
printf("%d ", arr[j]);
since the size of your array is stored in variable i. Also, the way you are calculating the size of your array is wrong which returns 1 everytime since the numerator and denominator are the same. Generally, it is sizeof(array)/sizeof(array[0]), and in this case, it too, would return 200, since the size of your declared array is 200, but since you increment your i every time at insertion, simply set your arr_size to be i.
int arr_size = i;
This line
int arr_size = sizeof(arr[i])/sizeof(arr[0]);
does not do what you are expecting. You are only dividing the size of two elements of the array, which are always the same size, thus always giving 1 as result. If you want to give the number of elements to the function, give it a variable that you used to count each number as you read them.
Also this:
for(i = 0; i < arr[i]; i++)
printf("%d ", arr[i]);
Is only printing numbers that are not larger than their indices. Again, your program is missing a variable to count the number of input values.