So what I am trying to do is create a random array of 5 elements, those elements should be filled with numbers from 1 to 6 and they shall not repeat, I can't tell where my logic is wrong.
void genNumber(int vet[]){
int max, i, j, atual;
srand(time(NULL));
max = 7;
for (i=0;i<5;i++){
vet[i] = rand() % max;
while(vet[i] == 0){
vet[i] = rand() % max;
}
for(j=0;j<i;j++){
atual = vet[j];
while((vet[i] == atual)||(vet[i] == 0)){
vet[i] = rand() % max;
atual = vet[j];
}
}
}
}
Update: Fixed
void genNumber(int vet[]){
int max, i, j;
srand(time(NULL));
max = 7;
for (i=0;i<5;i++){
vet[i] = rand() % (max-1) + 1;
for(j=0;j<i;j++){
while(vet[j] == vet[i]){
vet[i] = rand() % (max-1) + 1;
j = 0;
}
}
}
}
The logical flaw is in the way you produce a new random number when a duplicate is found.
Imagine you already have vel = {1,2,0,0,0,...} and are trying to find a number for vel[2]. If you randomly draw a 2, you'll find it's already there and draw again. But if you draw a 1 this time you won't notice, because you only compare to the last value seen, 2 in the example. So you'd get vel = {1,2,1,...}.
"Solution": Every time you draw a new random number you have to compare it against all numbers already in the list.
Another way of solving this is the one I tried to outline in the comments: You have to keep the information about which numbers are still valid for a draw somewhere. You can either use the "output" array for that as you're doing now, or you can use another store from which you "remove" an entry once it was drawn.
when generating a random number,you have to check if it is found in the array,so iterate over the array and check.if the number is found,generate another random number and check,until the new number is not found in the array.after that assign it.and repeat.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int check(int vet[],size_t size,int val);
void genNumber(int vet[],size_t size);
int main(void)
{
srand(time(NULL));
int vet[5];
size_t size = 5;
genNumber(vet,size);
for( size_t n = 0 ; n < size ; n++ )
{
printf("%d ",vet[n]);
}
}
void genNumber(int vet[],size_t size)
{
int num = rand() % 6 + 1;
vet[0] = num;
for( size_t n = 1 ; n < size ; n++ )
{
num = rand() % 6 + 1;
while( 1 )
{
if( check(vet,n,num) )
{
num = rand() % 6 + 1;
}
else
{
vet[n] = num;
break;
}
}
}
}
int check(int vet[],size_t size,int val)
{
for( size_t n = 0 ; n < size ; n++ )
{
if( vet[n] == val )
{
return 1;//FOUND
}
}
return 0;//NOT FOUND
}
As an alternative you can fill an array with a value from 1 to 6. Afterwards you can shuffle the array at random indicies.
#define MAX 6
void swap (int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
void genNumber(int *vet) {
int i;
for(i=0;i<MAX;i++) {
vet[i] = i+1;
}
for(i = MAX-1;i > 0;i--) {
// Pick a random index from 0 to i
int j = rand() % (i+1);
// Swap vet[i] with the element at random index
swap(&vet[i], &vet[j]);
}
}
Also you shouldn't call srand(time(NULL)); inside of the genNumber function if you intend to call it more than once, becaues otherwise it will give you the same numbers if you call it more than once a second.
Afterwards just use the first five elements from the array. So you would call it like this from main
int vet[MAX] = {0};
srand(time(NULL));
genNumber(vet);
for(i=0;i<5;i++) {
printf("after num %d\n", vet[i]);
}
Related
I need to write a program that allows user to enter an array of integers, finds the digit that appears most often in all entered numbers, and removes it from the elements of the array. If several digits appear the same number of times, the smallest of them should be deleted. If all digits of the element of the array are deleted, that element should become zero. In the end, such a modified array is printed.
Example of input and output:
Enter number of elements of the array: 5
Enter the array: 3833 8818 23 33 1288
After deleting, the array is: 8 8818 2 0 1288
Explanation: The numbers 3 and 8 appear the same number of times (6 times each), but 3 is less, so it was removed it from all members of the array. Element 33 consists exclusively of the digits 3, so that it becomes 0.
#include <stdio.h>
int main() {
int i,n,arr[100]; n;
printf("Enter number of elements of the array: ");
scanf("%d", &n);
printf("Enter the array: ");
for(i=0;i<n;i++) {
scanf("%d", &arr[i]);
}
return 0;
}
EDIT: I'm beginner to programming, and this task should be done using only knowledge learned so far in my course which is conditionals, loops, and arrays. This shouldn't be done with strings.
Divide the problem into separate tasks.
Write the code
In the code below I do not treat 0 as having digit 0. It is because it is not possible to remove 0 from 0. You can easily change this behaviour by changing while(){} loop to do{}while()
int removeDigit(int val, int digit)
{
int result = 0;
unsigned mul = 1;
int sign = val < 0 ? -1 : 1;
digit %= 10;
while(val)
{
int dg = abs(val % 10);
if(dg != digit)
{
result += dg * mul;
mul *= 10;
}
val /= 10;
}
return sign * result;
}
void countDigits(int val, size_t *freq)
{
while(val)
{
freq[abs(val % 10)]++;
val /= 10;
}
}
int findMostFrequent(const size_t *freq)
{
size_t max = 0;
for(size_t i = 1; i < 10; i++)
{
if(freq[i] > freq[max]) max = i;
}
return (int)max;
}
int main(void)
{
int table[20];
size_t freq[10] = {0,};
int mostfreq = 0;
srand(time(NULL));
for(size_t i = 0; i < 20; i++)
{
table[i] = rand();
printf("Table[%zu] = %d\n", i, table[i]);
countDigits(table[i], freq);
}
mostfreq = findMostFrequent(freq);
printf("Most frequent digit: %d\n", mostfreq);
for(size_t i = 0; i < 20; i++)
{
table[i] = removeDigit(table[i], mostfreq);
printf("Table[%zu] = %d\n", i, table[i]);
}
}
https://godbolt.org/z/PPj9s341b
I'm trying to write a program that will sort an array of 20 random numbers by the sums of their digits.
For example:
"5 > 11" because 5 > 1+1 (5 > 2).
I managed to sort the sums but is it possible to return to the original numbers or do it other way?
#include <stdio.h>
void sortujTab(int tab[], int size){
int sum,i;
for(int i=0;i<size;i++)
{
while(tab[i]>0){//sum as added digits of an integer
int p=tab[i]%10;
sum=sum+p;
tab[i]/=10;
}
tab[i]=sum;
sum=0;
}
for(int i=0;i<size;i++)//print of unsorted sums
{
printf("%d,",tab[i]);
}
printf("\n");
for(int i=0;i<size;i++)//sorting sums
for(int j=i+1;j<=size;j++)
{
if(tab[i]>tab[j]){
int temp=tab[j];
tab[j]=tab[i];
tab[i]=temp;
}
}
for(int i=0;i<20;i++)//print of sorted sums
{
printf("%d,",tab[i]);
}
}
int main()
{
int tab[20];
int size=sizeof(tab)/sizeof(*tab);
for(int i=0;i<=20;i++)
{
tab[i]=rand()%1000;// assamble the value
}
for(int i=0;i<20;i++)
{
printf("%d,",tab[i]);//print unsorted
}
printf("\n");
sortujTab(tab,size);
return 0;
}
There are two basic approach :
Create a function that return the sum for an integer, say sum(int a), then call it on comparison, so instead of tab[i] > tab [j] it becomes sum(tab[i]) > sum (tab[j])
Store the sum into a different array, compare with the new array, and on swapping, swap both the original and the new array
The first solution works well enough if the array is small and takes no extra memory, while the second solution didn't need to repeatedly calculate the sum. A caching approach is also possible with map but it's only worth it if there are enough identical numbers in the array.
Since your numbers are non-negative and less than 1000, you can encode the sum of the digits in the numbers itself. So, this formula will be true: encoded_number = original_number + 1000 * sum_of_the_digits. encoded_number/1000 will decode the sum of the digits, and encoded_number%1000 will decode the original number. Follow the modified code below. The numbers enclosed by parentheses in the output are original numbers. I've tried to modify minimally your code.
#include <stdio.h>
#include <stdlib.h>
void sortujTab(int tab[], int size)
{
for (int i = 0; i < size; i++) {
int sum = 0, n = tab[i];
while (n > 0) { //sum as added digits of an integer
int p = n % 10;
sum = sum + p;
n /= 10;
}
tab[i] += sum * 1000;
}
for (int i = 0; i < size; i++) { //print of unsorted sums
printf("%d%c", tab[i] / 1000, i < size - 1 ? ',' : '\n');
}
for (int i = 0; i < size; i++) { //sorting sums
for (int j = i + 1; j < size; j++) {
if (tab[i] / 1000 > tab[j] / 1000) {
int temp = tab[j];
tab[j] = tab[i];
tab[i] = temp;
}
}
}
for (int i = 0; i < size; i++) { //print of sorted sums
printf("%d(%d)%c", tab[i] / 1000, tab[i] % 1000, i < size - 1 ? ',' : '\n');
}
}
int main(void)
{
int tab[20];
int size = sizeof(tab) / sizeof(*tab);
for (int i = 0; i < size; i++) {
tab[i] = rand() % 1000; // assamble the value
}
for (int i = 0; i < size; i++) {
printf("%d%c", tab[i], i < size - 1 ? ',' : '\n'); //print unsorted
}
sortujTab(tab, size);
return 0;
}
If the range of numbers doesn't allow such an encoding, then you can declare a structure with two integer elements (one for the original number and one for the sum of its digits), allocate an array for size elements of this structure, and initialize and sort the array using the digit sums as the keys.
You can sort an array of indexes rather than the array with data.
#include <stdio.h>
//poor man's interpretation of sumofdigits() :-)
int sod(int n) {
switch (n) {
default: return 0;
case 5: return 5;
case 11: return 2;
case 1000: return 1;
case 9: return 9;
}
}
void sortbyindex(int *data, int *ndx, int size) {
//setup default indexes
for (int k = 0; k < size; k++) ndx[k] = k;
//sort the indexes
for (int lo = 0; lo < size; lo++) {
for (int hi = lo + 1; hi < size; hi++) {
if (sod(data[ndx[lo]]) > sod(data[ndx[hi]])) {
//swap indexes
int tmp = ndx[lo];
ndx[lo] = ndx[hi];
ndx[hi] = tmp;
}
}
}
}
int main(void) {
int data[4] = {5, 11, 1000, 9};
int ndx[sizeof data / sizeof *data];
sortbyindex(data, ndx, 4);
for (int k = 0; k < sizeof data / sizeof *data; k++) {
printf("%d\n", data[ndx[k]]);
}
return 0;
}
Goldbach's conjecture states that every even integer over 4 is the sum of two primes, I am writing a program in C to find these pairs. To do this it first finds all the primes less than a user given number. I have a for loop to iterate from 4 to the user given number and find the pairs within the loop body. When that loop gets to about around 40, suddenly jumps back down by about 30 and then continues to iterate up (with user input 50 it jumped from 38 to 9, with input 60 it jumped from 42 to 7). I can't figure out why this is happening. Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/types.h>
#include <unistd.h>
struct pair{
int a;
int b;
}pair_t;
int main(){
int N;
int numPrimes = 1;
int *primes = malloc(100*sizeof(int));
int isPrime = 1;
primes[0] = 2;
int timesRealloc = 0;
int availableSlots = 100;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
struct pair pairs[N/2 + 4];
int j = 0;
int i;
for (i = 3; i <= N; i+=2){
j = 0;
isPrime = 1;
while (primes[j] <= sqrt(i)) {
if (i%primes[j] == 0) {
isPrime = 0;
break;
}
j++;
}
if (isPrime == 1){
primes[numPrimes] = i;
numPrimes++;
}
if (availableSlots == numPrimes){
timesRealloc++;
availableSlots += 100;
primes = realloc(primes, availableSlots*sizeof(int));
}
}
printf("The largest prime I found was %d\n", primes[(numPrimes-1)]);
int k;
for (i=4; i<=N; i+=2){
printf("i is %d, N is %d\n", i, N);
if (i > N){ break; }
for (j=0; j<numPrimes; j++){
for (k=0; k<numPrimes; k++){
int sum = primes[j] + primes[k];
if(sum == i){
pairs[i].a = primes[j];
pairs[i].b = primes[k];
}
}
}
}
for (i=4; i<=N; i+=2){
printf("%d is the sum of %d and %d\n", i, pairs[i].a, pairs[i].b);
}
return 0;
}
You attempt to be space efficient by compressing the pairs array to just hold every other (even) number and start from 4 instead of zero. However, you miscalculate its size and then when you go to use it, you treat it like it hasn't been compressed and that there's a slot for every natural number.
The code suffers from having the prime array calculation in main() along with the other code, this is best separated out. And when it looks for pairs, it doesn't quit when it finds one, nor when it starts getting sums greater than the target. My rework below attempts to address all of these issues:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define INITIAL_SLOTS (100)
struct pair {
int a;
int b;
} pair_t;
int compute_primes(int limit, unsigned **primes, int size) {
int numPrimes = 0;
(*primes)[numPrimes++] = 2;
for (int i = 3; i <= limit; i += 2) {
bool isPrime = true;
for (int j = 0; (*primes)[j] <= i / (*primes)[j]; j++) {
if (i % (*primes)[j] == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
(*primes)[numPrimes++] = i;
}
if (numPrimes == size) {
size *= 2;
*primes = realloc(*primes, size * sizeof(unsigned));
}
}
return numPrimes;
}
int main() {
int N;
printf("Please enter the largest even number you want to find the Goldbach pair for: \n");
scanf("%d", &N);
unsigned *primes = calloc(INITIAL_SLOTS, sizeof(unsigned));
int numPrimes = compute_primes(N, &primes, INITIAL_SLOTS);
printf("The largest prime I found was %d\n", primes[numPrimes - 1]);
struct pair pairs[(N - 4) / 2 + 1]; // compressed data structure
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
bool found = false;
for (int j = 0; ! found && j < numPrimes; j++) {
for (int k = 0; ! found && k < numPrimes; k++) {
int sum = primes[j] + primes[k];
if (sum == i) {
pairs[offset].a = primes[j];
pairs[offset].b = primes[k];
found = true;
} else if (sum > i) {
break;
}
}
}
}
for (int i = 4; i <= N; i += 2) {
int offset = (i - 4) / 2; // compressed index
printf("%d is the sum of %d and %d\n", i, pairs[offset].a, pairs[offset].b);
}
free(primes);
return 0;
}
OUTPUT
> ./a.out
Please enter the largest even number you want to find the Goldbach pair for:
10000
The largest prime I found was 9973
4 is the sum of 2 and 2
6 is the sum of 3 and 3
8 is the sum of 3 and 5
10 is the sum of 3 and 7
12 is the sum of 5 and 7
14 is the sum of 3 and 11
...
9990 is the sum of 17 and 9973
9992 is the sum of 19 and 9973
9994 is the sum of 53 and 9941
9996 is the sum of 23 and 9973
9998 is the sum of 31 and 9967
10000 is the sum of 59 and 9941
>
I have to return which digit in a number occurs the most frequently ( though not how many times it occurs )
So far I can only get this, I don't know how to isolate the digit, only to show how many times each digit occurs.
#include <stdio.h>
int frequentDigit(int);
int main()
{
frequentDigit(123032333);
return 0;
}
int frequentDigit(int arg)
{
int tmp; int i; int myArr[9] = { 0 };
tmp = (arg < 0) ? -arg : arg;
do
{
myArr[tmp % 10]++;
tmp /= 10;
} while (tmp != 0);
for (i = 0; i < 9; i++) { printf("\nThere are %d occurances of digit %d", myArr[i], i); }
}
The array where you are storing the frequency of the digits, i.e myArr[]. Its suppose to hold the frequency of all the number from 0...9. And since there are 10 numbers, you would need an array of lenght 10.
int myArr[10];
Later, you need to traverse through the array once, checking for the max element, and saving the index accordingly, to find which number has occured most number of times.
To traverse, the for loop should go till 9
for (i = 0; i <= 9; i++)
Edited
As someone commented, you can find the max value while you are computing the frequencies itself.
int max = -1, max_num = -1;
do
{
myArr[tmp % 10]++;
if( myArr[tmp % 10] > max)
{
max = myArr[tmp % 10];
max_num = tmp % 10;
}
tmp /= 10;
} while (tmp != 0);
printf("%d", max_num);
Its simple. At the end of your code you have an array of frequencies, if you find the max of that you get the most common element
Just use a loop to find the max and print that:
int max = myArr[0]; // start with max = first element
int max_position=0;
for(int i = 1; i<9; i++)
{
if(myArr[i] > max){
max = myArr[i];
max_position=i;
}
}
printf("\The max is %d occuring %d times ", max_position, max_position)
int i,j,vec[15]={0};
srand (time(NULL));
for (i=0;i<15;i++){
vec[i]=rand() % 25+1;
for (j=0;j<15;j++){
if (i!=j){
while(vec[i]==vec[j]){
vec[i]=rand() % 25+1;
}
}
}
printf("%d\n",vec[i]);
}
return 0;
}
the code still gives me repeated numbers
EXAMPLE:
24
3
7
20
18
10
12
17
9
7
4
25
13
15
21
I cant figure out what to do with it
You have your loops mixed up. The logic is: Generate a random number until you have found one that isn't in the list.
The way you do it, you generate a new number inside the checking loop. But that doesn't work. Say you're generating the 4th number and find it is equal to the third. Then you generate a new one which might well be equal to any you have already checked against.
You also check uninitialised elements when j > i. Your inner loop should only run up to i.
So:
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
int main()
{
int vec[15] = { 0 };
int i, j;
srand(time(NULL));
for (i = 0; i < 15; i++) {
int okay = 0;
while (!okay) {
vec[i] = rand() % 25 + 1;
okay = 1;
for (j = 0; j < i; j++) {
if (vec[i] == vec[j]) okay = 0;
}
}
printf("%d\n", vec[i]);
}
return 0;
}
That still looks a bit awkward with that okay variable. In my opinion, checking for duplicates should be a separate function:
int contains(int arr[], int n, int x)
{
while (n--) {
if (arr[n] == x) return 1;
}
return 0;
}
int main()
{
// snip ...
for (i = 0; i < 15; i++) {
do {
vec[i] = rand() % 25 + 1;
} while (contains(vec, i, vec[i]));
printf("%d\n", vec[i]);
}
// snip ...
}
In your case the range of possible numbers isn't mich bigger than the number of array elements. You could also create an ordered array {1, 2, 3, ..., 25}, then shuffle it and use only the first 15 elements.
Reset j in the while loop:
for (j=0;j<i;j++){ //Use j<i
if (i!=j){
while(vec[i]==vec[j]){
vec[i]=rand() % 25+1;
j=-1;//-1 because in the next iteration,j will start from 0
}
}
}
Are you actually trying to shuffle the numbers, rather than fill the array with randoms? (It looks like you want an array with numbers from 1 to 25, but in random order.) rand() can give you duplicate numbers (they're random, after all!)
Try this:
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <time.h>
int
main( int argc, char **argv )
{
int i, vec[25];
for (i = 0; i < 25; ++i) vec[i] = i + 1;
/* Shuffle entries */
srand( time( 0 ) );
for (i = 0; i < 1000; ++i) {
int a = rand( ) % 25;
int b = rand( ) % 25;
if (a != b) {
int tmp = vec[a];
vec[a] = vec[b];
vec[b] = tmp;
}
}
/* Print shuffled array */
for (i = 0; i < 25; ++i) printf( "%d: %d\n", i, vec[i] );
return 0;
}
#include<stdio.h>
#include<stdlib.h>
int inArray(int, int, int*);
int main()
{
int i,j,vec[15]={0};
int temp;
srand (time(NULL));
for (i=0;i<15;i++){
temp =rand() % 25+1;
while(inArray(i+1,temp, vec) == 1){
temp = rand() % 25+1;
}
vec[i] = temp;
printf("VECT[%d] \t= %d\n",i,vec[i]);
}
return 0;
}
int inArray(int count, int input, int* array){
int i = 0;
for(i=0; i<count; i++){
if(input == array[i]){
return 1;
}
}
return 0;
}
Gave an output:
VECT[0] = 24
VECT[1] = 19
VECT[2] = 1
VECT[3] = 25
VECT[4] = 22
VECT[5] = 18
VECT[6] = 7
VECT[7] = 8
VECT[8] = 12
VECT[9] = 21
VECT[10] = 11
VECT[11] = 6
VECT[12] = 23
VECT[13] = 20
VECT[14] = 15
The checking was off, you would change and not break allowing it to be changed to a previous value.
You can use an array
int randNumbers[25]; // fill it starting 0 to 25 then
randomize the number in a range between 0 and 25 after swap the number in the randomized index with the last number in your array
randomize 0 to 23
and so on....
int main(int argc, char **argv) {
static const int size = 25;
int numbers[size];
for( int i = 0; i < size; i++ ){
numbers[i] = i;
}
srand (time(NULL));
for( int i = 0; i < size; i++ ){
int rIndex = rand()%(size - i);
int rNum = numbers[rIndex];
numbers[rIndex] = numbers[size-i];
printf("%d ", rNum);
}
return 0;
}
O(n) complexity...