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.
Related
How can i write a program that lists all sexy prime pairs that exist in n numbers.
For example if n = 10 the output should be (5, 11) and (7, 13)
My idea was to generate all primes within n and then add 6 to each and check if the i + 6 is a prime. But it doesnt work, there's no output and the program ends.
#include <stdio.h>
int main() {
int i, j, n, k, isprime = 1, prime2, flag = 0;
scanf("%d", &n);
for (i = 3; i <= n; i++){
for (j = 2; j <= i; j++){
if (i % j == 0)
break;
}
if (i == j){
prime2 = i + 6;
for (k = 3; k <= prime2; k++){
if (prime2 % k == 0){
flag++;
break;
}
}
if (flag == 0){
printf("%d %d\n", i, prime2);
}
}
}
return 0;
}
Any ideas of what im doing wrong or any tips on how to solve it? (with loops only)
As there're a lot of resources about finding a prime number, I'm not going to discuss that. Rather I'll try to point out the bug in your code.
First problem:
for (k = 3; k <= prime2; k++)
Here you need to run the loop till prime2 - 1. Also you should start checking from 2 rather than 3, just like you did previously. That means,
for (k = 2; k < prime2; k++)
or
for (k = 2; k <= prime2 - 1; k++)
Reason: when k = prime2, prime2 % k will be 0. For finding out whether a number is prime we don't need to check if that number is divisible by 1 and that number itself.
Note: Now you might think why the first prime number loop for (j = 2; j <= i; j++) is working .
It's working because you've given an additional condition if (i == j) after it.
Second problem:
You need to declare the flag variable within the first loop.
for (i = 2; i <= n; i++)
{
int flag = 0;
.... (rest of the code)
....
}
Reason: Basically with the flag value, you're trying to find out whether prime2 is a prime number.
Every time you'll get a prime number from the first loop, you'll have a new value of prime2. In your code, once you're incrementing the value of flag, you're never resetting the flag value.
That's why once your code detects a prime2 which is not a prime, it'll never detect the second prime number again (prime2 which is actually prime).
Overall code:
#include <stdio.h>
int main()
{
int i, j, n, k, isprime = 1, prime2;
scanf("%d", &n);
for (i = 3; i <= n; i++)
{
int flag = 0; // changing point
for (j = 2; j <= i; j++)
{
if (i % j == 0)
break;
}
if (i == j)
{
prime2 = i + 6;
for (k = 2; k < prime2; k++) // changing point
{
if (prime2 % k == 0)
{
flag++;
break;
}
}
if (flag == 0)
{
printf("%d %d\n", i, prime2);
}
}
}
return 0;
}
Few resources to know more about finding out prime numbers:
Prime Numbers
C Program to Check Whether a Number is Prime or not
Sieve of Eratosthenes
You can use Sieve to speed up the program. It can generate all pairs in O(N log N) time. Here's the Algorithm.
Now, you have a boolean array, is_prime where is_prime[i] is true if i is a prime, false otherwise.
Now, iterate from i = 1 to i = N and check if is_prime[i] && is_prime[i + 6], if the condition is true, output the pair.
I'm trying to sort a set of numbers like this:
A[]={3,6,7,2,9,1,2,7,2}
A[]={3,6,7,2,2,2,9,1,7}
So I've made this:
void sort_min(int* point, int size_array, int min_n){
int i = 0;
int j = 0;
int k = 0;
while(point[i] != min_n){
i++;
}
j = i+1;
while(point[j] != min_n){
j++;
}
k = j;
for (j-1; j > i; j--){
point[j] = point[j-1];
}
point[j] = min_n;
j = k+1;
}
Like you can notice I've never used the int size_array cause I don't know the way to match an iterative function like a for or a while (That's the question. How to solve it?). I've done that, of course, but I've got a Segmentation fault like an answer.
The main concept is looking for a number int min_n and up to that point sort that number at each occurrence in the array.
Thanks for all.
You need use size_array like below, if you are asking about this.
You need compare i and j with size_array in while.
while (i < size_array && point[i] != min_n) {
i++
}
Need check value of i, j after while. They may be larger or equal to size_array.
while (i < size_array && point[i] != min_n) {
i++
}
// I guess when you don't find min_n, function can just return.
if (i >= size_array)
return;
j = i+1;
while(j < size_array && point[j] != min_n){ // Also need check j's value.
j++;
}
// Also guess when can't find the second min_n position, function can return.
if (j >= size_array)
return;
k = j
for (; j > i; j--) // No need j-1.
point[j] = point[j-1];
// This is useless. When code come here, j == i and point[i] == min_n;
point[j] = min_n;
j = k+1;
Try the Insertion Sort with the code as follows..
void sort_min(int point[], int size_array)
{
int i, key, j;
for (i = 1; i < size_array; i++)
{
key = point[i];
j = i-1;
/* Move elements of arr[0..i-1], that are
greater than key, to one position ahead
of their current position */
while (j >= 0 && point[j] > key)
{
point[j+1] = point[j];
j = j-1;
}
point[j+1] = key;
}
}
#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.
The purpose of the program is to create a random list of 1000 numbers in an array, sort that array, then find the greatest set of numbers within (x, x+50). The program successfully generates and sorts the numbers within the array, but crashes when the (i, j) set finding algorithm starts. The program generates no errors on compiling, and I'm sure the error is simple, but for the life of me I can't find the issue. Thanks in advance you amazing people!
int main( ){
int a, b, temp, i, j, x, y, tempTotal, arrayStartMax;
int finalTotal = 0;
int *info[ARRAY_FULL];
for (i=0; i<ARRAY_FULL; i++){
info[i]=(int*)malloc(sizeof(int));
*info[i]=rand()%1000;
}
for (a = 0; a < ARRAY_FULL; ++a){
for (b = a + 1; b < ARRAY_FULL; ++b){
if (*info[a] > *info[b]){
temp = *info[a];
*info[a] = *info[b];
*info[b] = temp;
}
}
}
for (i=0; i<ARRAY_FULL; i++){
printf("%d\n", *info[i]);
}
for (i = 0; i <= ARRAY_HALF; i++){
x = *info[i];
y = x+ARRAY_HALF;
tempTotal = 0;
for (j = i; j < i+ARRAY_HALF; i++){
if (*info[j] >= x || *info[j] <= y) {
tempTotal++;
}
if (tempTotal > finalTotal) {
arrayStartMax = *info[i];
finalTotal = tempTotal;
}
}
}
printf("Interval should start at %d for maximum numbers in a set.", arrayStartMax);
}
For the purpose of this program I would like to mention that ARRAY_FULL = 100 and ARRAY_HALF = 50.
Your code is throwing segfault because you're walking i out of bounds in this for loop.
for (j = i; j < i+ARRAY_HALF; i++){
if (*info[j] >= x || *info[j] <= y) {
tempTotal++;
}
if (tempTotal > finalTotal) {
arrayStartMax = *info[i];
finalTotal = tempTotal;
}
You set j = i then increment i prior to the comparison. So j will always be less than i.
Limit i in the comparison section of the for loop and it won't segfault.
I don't think the comparison is doing what you want, but you should be able to find your way home from here.
I have an array that I am trying to remove values from, however I run into 2 problems when I try to remove from the array:
1.) When I try to remove from an array that is full, I get a * Stack Smashing Detected * error
2.) when my array has multiple duplicates, I only want to remove the first instance, however it seems to remove a random number of instances.
The code I have for the remove function is as follows:
int Remove(int* array, int arrayLen, int removeNum)
{
int i, j;
for (i = 0; i < arrayLen; i++)
{
if (array[i] == removeNum){
for(j = i; j < arrayLen; j++){
array[j] = array[j+1];
}
arrayLen--;
}
}
return arrayLen;
}
Where removeNum is user selected in main().
I see two issues:
You're reading past the end of the list because you're inner loop goes too far. (At j = arrayLen - 1, j + 1 = arrayLen, which is past the end of the array.)
You say you only want to remove the first found element, but your code continues past that point. Adding an early return (or break) fixes that problem.
Here's a fixed version of your code:
int remove(int* array, int arrayLen, int removeNum)
{
int i, j;
for (i = 0; i < arrayLen; i++)
{
if (array[i] == removeNum) {
// Stop j at arrayLen - 1, which is the new end of the array
for(j = i; j < arrayLen - 1; j++) {
array[j] = array[j + 1];
}
// Return the new size
return arrayLen - 1;
}
}
// If we didn't remove anything, return the original size
return arrayLen;
}