I'm trying to write a code that could print something like this
-xv
-xvv
-xvvv
-xvvvv
-xvvvvv
-xxv
-xxvv
-xxvvv
-xxvvvv
-xxvvvvv
-xxxv
-xxxvv
-xxxvvv
-xxxvvvv
-xxxvvvvv
The extra spacing between the 3 "groups" are just for clarity's sake. The maximum number of '-' is 1, 'x' is 3, 'v' is 5, and the number of each symbol increments.
To draw this diagram, I have the following code
for (k = 1 ; k <= num_dash ; k++)
{
for (i = 1 ; i <= num_x ; i++)
{
for (j = 1 ; j <= num_v ; j+++)
{
for (k1 = 0 ; k1 < k ; k++)
printf("-");
for (i1 = 0 ; i1 < i ; i++)
printf("x");
for (j1 = 0 ; j1 < j ; j++)
printf("v");
printf("\n");
}
}
}
This is when I know there are 3 different kinds of symbols. Is it possible to do the same if the number of symbols are only known at runtime? For example, what if I want the same program to also be able to print
xv
xvv
xvvv
xvvvv
xvvvvv
xxv
xxvv
xxvvv
xxvvvv
xxvvvvv
xxxv
xxxvv
xxxvvv
xxxvvvv
xxxvvvvv
In this case, my code would only be
for (i = 1 ; i <= num_x ; i++)
{
for (j = 1 ; j <= num_v ; j+++)
{
for (i1 = 0 ; i1 < i ; i++)
printf("x");
for (j1 = 0 ; j1 < j ; j++)
printf("v");
printf("\n");
}
}
And there are only 2 for-loops instead of 3. Can I write my code such that the number of for-loops varies?
Use functions instead of innermost for loops and pass the characters and number of iterations as parameters to that function.
E.g.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void print_aux(char **symbols, int *times, int len, int pos, char *acc){
int i;
char *p;
if(pos == len){
printf("%s\n", acc);
return;
}
p = malloc(sizeof(char)*strlen(symbols[pos])*times[pos]+strlen(acc)+1);
for(i=0;i<times[pos];++i){
int j;
*p='\0';
strcpy(p, acc);
for(j=0;j<=i;++j){
strcat(p, symbols[pos]);
}
print_aux(symbols, times, len, pos + 1, p);
}
if(pos + 1 == len)// when last symbol
printf("\n",pos);
free(p);
}
//wrap function
void print(char **symbols, int *times, int len){
print_aux(symbols, times, len, 0, "");
}
int main() {
int i,n;
char **symbols;
int *times;
fprintf(stderr,"number of symbols :");
scanf("%d", &n);
symbols=(char**)malloc(sizeof(char*)*n);
times = (int*)malloc(sizeof(int)*n);
for(i=0;i<n;++i){
char wk[128];
fprintf(stderr,"input symbol [%d]:", i+1);
scanf(" %s", wk);
symbols[i] = strdup(wk);
fprintf(stderr,"maximum number of \"%s\":",wk);
scanf(" %d", ×[i]);
}
print(symbols, times, n);
{ //release the allocated area
for(i=0;i<n;++i){
free(symbols[i]);
}
free(symbols);
free(times);
}
return 0;
}
Related
I need to do the following pattern using do-while, while or for.
I tried the following code but it prints the pattern only 1-5
I also tried to alter the n being 10 but then the spacing goes nuts.
#include <stdio.h>
int main(void)
{
int n = 5, i, j, num = 1, gap;
gap = n - 1;
for ( j = 1 ; j <= n ; j++ )
{
num = j;
for ( i = 1 ; i <= gap ; i++ )
printf(" ");
gap --;
for ( i = 1 ; i <= j ; i++ )
{
printf("%d", num);
num++;
}
num--;
num--;
for ( i = 1 ; i < j ; i++)
{
printf("%d", num);
num--;
}
printf("\n");
}
return 0;
}
Try replacing both occurences of:
printf("%d", num);
with
printf("%d", num % 10);
Now only the last digit will be shown.
After the change, for n set to 10 the program produces:
1
232
34543
4567654
567898765
67890109876
7890123210987
890123454321098
90123456765432109
0123456789876543210
Sample Input
2
2
5
Sample Output
0 1 1
0 1 1 2 1 2
I already knew how to transfer an integer to its binary and count how many 1 in its binary.
But my code only can input one integer each time. I want it to input many numbers, like the sample input and sample output. To make question more easy to understand, so I drew a picture. Thank you!!
Thanks all of you!! But I need some time to understand those code and my English is pretty basic, so I couldn't reply you guys soon. But I will understand and reply you as soon as possible!!thank you :D
#include <stdio.h>
int main()
{
int n,cnt=0,m;
scanf("%d",&n);
while(n>0){
m=n%2;
if(m==1){
cnt++;
}
n/=2;
}
printf("%d",cnt);
return 0;
}
In practice, the process can be simplified and accelerated by noticing that if we know the result for i < 2^p, then, for all values in [2^p, 2^(p+1)-1], we have
count(j) = 1 + count(j-2^p)
This method is useful here as we have to provide the result for all values less or equal to n.
Moreover, in order to avoid performing the same calculation at different times, we first calculate the maximum n value, and perform the calculation for this maximum value.
#include <stdio.h>
int main() {
int t;
int check = scanf("%d", &t);
if (check != 1) return 1;
int nn[t];
for (int i = 0; i < t; ++i) {
int check = scanf("%d", &nn[i]);
if (check != 1) return 1;
}
int nmax = 0;
for (int i = 0; i < t; ++i) {
if (nn[i] > nmax) nmax = nn[i];
}
int count[nmax+1];
count[0] = 0;
count[1] = 1;
int pow2 = 1;
do {
pow2 *= 2;
for (int j = pow2; (j < 2*pow2) && (j <= nmax); j++) {
count[j] = 1 + count[j-pow2];
}
} while (pow2 <= nmax+1);
for (int i = 0; i < t; ++i) {
for (int j = 0; j <= nn[i]; ++j) {
printf ("%d ", count[j]);
}
printf ("\n");
}
return 0;
}
Let the number of measured values be t. Now if you want to take t number of integers as input, you have to run a loop t times and perform necessary tasks (like taking value of n, calculating cnt etc) inside that loop. Check the following code snippet:
int main()
{
int n,cnt=0,m;
int numberOfMeasuredValues;
scanf("%d", &numberOfMeasuredValues);
while(numberOfMeasuredValues > 0){
cnt = 0;
scanf("%d",&n);
// perform necessary calculations
printf("%d\n",cnt);
numberOfMeasuredValues--;
}
return 0;
}
In the code numberOfMeasuredValues is the t I've used in the explanation.
You can pre-allocate number of cases and work on them with outer and inner loops. I've added the code with comments below.
#include <stdio.h>
int main(void)
{
int cases; // Number of cases
scanf("%d", &cases);
int* nums = malloc(cases * sizeof *nums); // Allocate enough space.
/* Take the inputs. */
for (int i = 0; i < cases; i++)
{
scanf("%d", (nums + i));
}
int cnt = 0; // Counter for number of 1s.
// Outer loop for going through cases.
for (int i = 0; i < cases; i++)
{
// Inner loop for couting towards inputted number.
for (int j = 0; j < *(nums + i) + 1; j++)
{
/*
* You can Implement this part by yourself if you wish.
*/
for (int k = 0; k < 32; k++)
{
if ((j >> (31 - k)) & 0x01) cnt++; // Increment counter if digit is 1.
}
printf("%d ", cnt);
cnt = 0;
}
printf("\n");
}
return 0;
}
Here's the following code to your problem
#include<iostream>
using namespace std;
int main()
{
int tcs;
cin >> tcs;
while (tcs--) {
int n;
cin >> n;
for (int i = 0 ; i <= n ; i++) {
cout << __builtin_popcount(i) << " ";
}
cout << "\n";
}
return 0;
}
My aim is to generate a histogram for repeated numbers. The code works well until the frequency is bigger than 2.
I think I know what is wrong with the code (line 9) but I cannot find an algorithm to solve it. The problem that I have is when it writes the histogram, it separates and then gathers it again.
My Input:
5
5 6 6 6 7
Output:
6:2 6:2 6:3
but the output I need is
6:3
I kind of see the problem but I couldn't solve it.
#include <stdio.h>
int main(){
int array[25];
int i, j, num, count = 1;
scanf("%d", &num);
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
for (j = 0; j < i ; j++) {
if (array [i] == array[j]) {
count++;
printf("%d:%d ", array[i], count);
}
}
array [i] = array[j];
count = 1;
}
return 0;
}
You are trying to count occurrences before all units have been accepted, which is not possible unless you maintain a separate counter for each value, which in turn is not practical if there is no restriction on the input value range or the range is large.
You need to have obtained all values before you can report any counts. Then for each value in the array, test if the value has occurred earlier, and if not, iterate the whole array to count occurrences:
#include <stdio.h>
#include <stdbool.h>
int main()
{
// Get number of values
int num = 0 ;
scanf("%d", &num);
// Get all values
int array[25];
for( int i = 0; i < num; i++)
{
scanf("%d", &array[i]);
}
// For each value in array...
for( int i = 0; i < num ; i++)
{
// Check value not already counted
bool counted = false ;
for( int j = 0; !counted && j < i; j++ )
{
counted = array[j] == array[i] ;
}
// If current value has not previously been counted...
if( !counted )
{
// Count occurnaces
int count = 0 ;
for( int j = 0; j < num; j++ )
{
if( array[j] == array[i] )
{
count++ ;
}
}
// Report
printf("%d:%d ", array[i], count);
}
}
return 0;
}
For your example input, the result is:
5
5 6 6 6 7
5:1 6:3 7:1
It is possible to merge the two inner loops performing the counted and count evaluation:
// Count occurrences of current value,
bool counted = false ;
int count = 0 ;
for( int j = 0; !counted && j < num; j++ )
{
if( array[j] == array[i] )
{
count++;
// Discard count if value occurs earlier - already counted
counted = j < i ;
}
}
// If current value has not previously been counted...
if( !counted )
{
// Report
printf("%d:%d ", array[i], count);
}
}
I should make new array out of existing one (ex. 1 0 4 5 4 3 1) so that the new one contains digits already in existing array and the number of their appearances.
So, the new one would look like this: 1 2 0 1 4 2 5 1 3 1 (1 appears 2 times, 0 appears 1 time.... 3 appears 1 time; the order in which they appear in first array should be kept in new one also); I know how to count no. of times a value appears in an array, but how do I insert the no.of appearances? (C language)
#include <stdio.h>
#define max 100
int main() {
int b, n, s, i, a[max], j, k;
printf("Enter the number of array elements:\n");
scanf("%d", &n);
if ((n > max) || (n <= 0)) exit();
printf("Enter the array:\n");
for (i = 0; i < n; i++)
scanf("%d", a[i]);
for (i = 0; i < n; i++) {
for (j = i + 1; j < n;) {
if (a[j] == a[i]) {
for (k = j; k < n; k++) {
a[k] = a[k + 1];
}}}}
//in the last 5 rows i've tried to compare elements, and if they are same, to increment the counter, and I've stopped here since I realised I don't know how to do that for every digit/integer that appears in array//
If you know that the existing array consists of digits between 0 and 9, then you can use the index of the array to indicate the value that you are incrementing.
int in[12] = {1,5,2,5,6,5,3,2,1,5,6,3};
int out[10] = {0,0,0,0,0,0,0,0,0,0};
for (int i = 0; i < 12; ++i)
{
++out[ in[i] ];
}
If you provide any code snippet, its easy for the community to help you.
Try this, even you optimize the no.of loops :)
#include <stdio.h>
void func(int in[], int in_length, int *out[], int *out_length) {
int temp[10] = {0}, i = 0, j = 0, value;
//scan the input
for(i=0; i< in_length; ++i) {
value = in[i];
if(value >= 0 && value <= 9) { //hope all the values are single digits
temp[value]++;
}
}
// Find no.of unique digits
int unique_digits = 0;
for(i = 0; i < 10; ++i) {
if(temp[i] > 0)
unique_digits++;
}
// Allocate memory for output
*out_length = 2 * unique_digits ;
printf("digits: %d out_length: %d \n",unique_digits, *out_length );
*out = malloc(2 * unique_digits * sizeof(int));
//Fill the output
for(i = 0, j = 0; i<in_length && j < *out_length; ++i) {
//printf("\n i:%d, j:%d val:%d cout:%d ", i, j, in[i], temp[in[i]] );
if(temp[in[i]] > 0 ) {
(*out)[j] = in[i];
(*out)[j+1] = temp[in[i]];
temp[in[i]] = 0; //Reset the occurrences of this digit, as we already pushed this digit into output
j += 2;
}
}
}
int main(void) {
int input[100] = {1, 0, 4, 5, 4, 3, 1};
int *output = NULL, output_length = 0, i = 0;
func(input, 7, &output, &output_length );
for(i=0; i < output_length; i+=2) {
printf("\n %d : %d ", output[i], output[i+1]);
}
return 0;
}
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
#include <stddef.h>
void insertionSort(int ar_size, int* ar) {
int i, j;
int temp = 0;
for (i = 1; i < ar_size; i++) {
j = i;
while (j > 0 && ar[i - 1] > ar[i]) {
temp = ar[i - 1];
ar[i - 1] = ar[i];
ar[i] = temp;
j--;
}
}
for (j = 0; j < ar_size; j++) {
printf("%d", ar[j]);
printf(" ");
}
}
int main(void) {
int _ar_size;
scanf("%d", &_ar_size);
int _ar[_ar_size], _ar_i;
for (_ar_i = 0; _ar_i < _ar_size; _ar_i++) {
scanf("%d", &_ar[_ar_i]);
}
insertionSort(_ar_size, _ar);
return 0;
}
I have been trying to look for the error. I cannot see any. What is wrong with this code?
For input as 6 and 4 1 3 5 6 2 , it gives output as 1 3 4 5 2 6. There is one less iteration of the loop but I cannot see why? Please help. Thanks.
You are using index i instead of index j in the internal loop of the function.
while(j>0 && ar[i-1]>ar[i])
{
temp = ar[i-1];
ar[i-1] = ar[i];
ar[i] = temp;
j--;
}
Here everywhere index j has to be used.
Also it is a bad idea that the function also outputs the sorted array. It should do only sorting.
The other bad idea is to use identifiers that start with undescores.
It is better when the first parameter is an array and the second parameter is the size of the array.
The code could look the following way
#include <stdio.h>
void InsertionSort( int *a, int n )
{
int i;
for ( i = 1; i < n; i++ )
{
int j = i;
while ( j > 0 && a[j-1] > a[j] )
{
int tmp = a[j-1];
a[j-1] = a[j];
a[j] = tmp;
--j;
}
}
}
int main(void)
{
int size;
scanf( "%d", &size );
int a[size];
int i;
for ( i = 0; i < size; i++ ) scanf( "%d", &a[i] );
for ( i = 0; i < size; i++ )
{
printf( "%d ", a[i] );
}
puts( "" );
InsertionSort( a, size );
for ( i = 0; i < size; i++ )
{
printf( "%d ", a[i] );
}
puts( "" );
return 0;
}
If the input is
10
2 7 5 4 9 1 4 8 3 5
then output is
2 7 5 4 9 1 4 8 3 5
1 2 3 4 4 5 5 7 8 9
You seem to be using the wrong iterator when pushing the value to the front.
while(j>0 && ar[j-1]>ar[j]) {
temp = ar[j-1];
ar[j-1] = ar[j];
ar[j] = temp;
j--;
}
Try performing a dry run on a piece of paper, it's easy to spot the problem.
for (i = 1; i < ar_size; i++) {
j = i;
while (j > 0 && ar[i-1] > ar[i]) { // problem begins here
temp = ar[i-1];
ar[i-1] = ar[i];
ar[i] = temp;
j--;
}
}
i does not change in the inner loop.
At some point your program swaps 6 with 2 and your sample input list becomes 1 3 4 5 2 6, a[i-1] is 2 and a[i] is 6 and because of ar[i-1] > ar[i] condition the program's flow does not go into the inner loop.
Try this fix:
for (i = 1; i < ar_size; i++) {
j = i;
while (j > 0 && ar[j-1] > ar[j]) {
temp = ar[j-1];
ar[j-1] = ar[j];
ar[j] = temp;
j--;
}
}
The problem with your insertion-sort is you have your j indexes replaced with i indexes in the following code:
while(j>0 && ar[j-1]>ar[j])
{
temp = ar[j];
ar[j] = ar[j-1];
ar[j-1] = temp;
j--;
}
Just a note, when requesting input, it is good practice to print a brief statement describing the input expected. (yes, quick a dirty testing is an exception) It makes if much easier to avoid mistakes with something like:
printf ("enter array size: ");
scanf ("%d", &_ar_size);
int _ar[_ar_size], _ar_i;
for (_ar_i = 0; _ar_i < _ar_size; _ar_i++) {
printf ("enter array element[%d] : ", _ar_i);
scanf ("%d", &_ar[_ar_i]);
}