I'm trying to solve a problem, The objective is to print the last step can be achieved, the inputs are N as testcases, and A as the integers of the testcases. For example :
N = 10
A = 1 2 1 1 2 3 4 1 2 3 <- 10 integers
So, this is what I have done.
#include <stdio.h>
int main()
{
int N,i,j;
scanf("%d",&N);
int A[N+1];
for(int i=0;i<N;i++)
{
scanf("%d",&A[i]);
}
A[N] = 1;
for(int i=0;i<N;i++)
{
if(A[i] > A[i+1])
{
printf("%d-",A[i]);
}
else if(A[i] == A[i+1])
{
printf("%d-",A[i]);
}
}
printf("\n");
return 0;
}
The expected output is :
2-1-4-3
but the output I get is :
2-1-4-3-
How do I exclude the minus symbol after the last output? Thanks in advance.
Personally, my preference is to print the separator character before. The simple reason is that generally this happens in loops that begin at zero, so I can do a zero-test as follows:
for(int i = 0; i < N; i++)
{
if (i > 0)
putc('-', stdout);
printf("%d", A[i]);
}
BUT, in your case, your loop won't always print a value. So you actually need to be smarter. There are many ways to achieve this, but for simplicity, why not just use a flag:
int has_output = 0;
for(int i = 0; i < N; i++)
{
if(A[i] >= A[i+1])
{
if (has_output)
putc('-', stdout);
else
has_output = 1;
printf("%d", A[i]);
}
}
Notice that the preference is still to print the separator just in time. In other words, only when you determine that you need to print something.
Going a bit more crazy:
const char* fmt[2] = { "%d", "-%d" };
int has_output = 0;
for(int i = 0; i < N; i++)
{
if(A[i] >= A[i+1])
{
printf(fmt[has_output], A[i]);
has_output = 1;
}
}
#include<stdio.h>
int main()
{
int N,i,j;
scanf("%d",&N);
int A[N+1];
for(int i=0;i<N;i++)
{
scanf("%d",&A[i]);
}
A[N] = 1;
int flag = 0; //added a new variable here
for(int i=0;i<N;i++)
{
if(A[i] >= A[i+1])
{
if(flag == 0){ //check if its printing for the first time
printf("%d",A[i]); // then print without dash
flag = 1; //change the flag so that this block never executes again
}else{
printf("-%d",A[i] ); //else print a dash and then the number
}
}
}
printf("\n");
return 0;
}
As "-" is a separator only used after the first printed value, consider changing the separator.
const char *separator = "";
for(int i=0;i<N;i++) {
if(A[i] >= A[i+1]) {
printf("%s%d", separator, A[i]);
separator = "-";
}
}
printf("\n");
Related
I am having trouble achieving the wanted results. The program should ask for 20 inputs and then go over each to see if they appear more than once. Then only print out those that appeared once.
However currently my program prints out random numbers that are not inputted.
For example:
array = {10,10,11,12,10,10,10.....,10} should return 11 and 12
#include <stdio.h>
void main() {
int count, size=20, array[size], newArr[size];
int number=0;
for(count = 0; count < size; count++) {
// Ask user for input until 20 correct inputs.
printf("\nAnna %d. luku > ", count+1);
scanf("%d", &number);
if( (number > 100) || (number < 10) ) {
while(1) {
number = 0;
printf("Ei kelpaa.\n");//"Is not valid"
printf("Yrita uudelleen > ");//"Try again >"
scanf("%d", &number);
if ( (number <= 100) && (number >= 10) ) {
break;
}
}
}
array[count] = number;
}
for(int i=0; i < size; i++) {
for(int j=0; j<size; j++){
if(array[i] == array[j]){
size--;
break;
} else {
// if not duplicate add to the new array
newArr[i] == array[j];
}
}
}
// print out all the elements of the new array
for(int k=0; k<size; k++) {
printf("%d\n", newArr[k]);
}
}
You don't need the newArr here, or the separate output loop. Only keep a count that you reset to zero at the beginning of the outer loop, and increase in the inner loop if you find a duplicate.
Once the inner loop is finished, and the counter is 1 then you don't have any duplicates and you print the value.
In code perhaps something like:
for (unsigned i = 0; i < size; ++i)
{
unsigned counter = 0;
for (unsigned j = 0; j < size; ++j)
{
if (array[i] == array[j])
{
++counter;
}
}
if (counter == 1)
{
printf("%d\n", array[i]);
}
}
Note that the above is a pretty naive and brute-force way to deal with it, and that it will not perform very well for larger array sizes.
Then one could implement a hash-table, where the value is the key, and the count is the data.
Each time you read a value you increase the data for that value.
Once done iterate over the map and print all values whose data (counter) is 1.
Use functions!!
Use proper types for indexes (size_t).
void printdistinct(const int *arr, size_t size)
{
int dist;
for(size_t s = 0; s < size; s++)
{
int val = arr[s];
dist = 1;
for(size_t n = 0; n < size; n++)
{
if(s != n)
if(val == arr[n]) {dist = 0; break;}
}
if(dist) printf("%d ", val);
}
printf("\n");
}
int main(void)
{
int test[] = {10,10,11,12,10,10,10,10};
printdistinct(test, sizeof(test)/sizeof(test[0]));
fflush(stdout);
}
https://godbolt.org/z/5bKfdn9Wv
This is how I did it and it should work for your:
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <stdarg.h>
void printdistinct(const int *muis, size_t size);
int main()
{
int loop=20,i,muis[20],monesko=0;
for(i=0; i<loop; i++){
monesko++;
printf ("Anna %d. luku: \n",monesko);
scanf("%d", &muis[i]);
if (muis[i]<10 || muis[i]>100){
printf("Ei kelpaa!\n");
muis[i] = muis[i + 1];
printf("YRITÄ UUDELLEEN:\n ");
scanf("%d", &muis[i]);
}
}
printdistinct(muis, sizeof(muis)/sizeof(muis[0]));
fflush(stdout);
return 0;
}
void printdistinct(const int *muis, size_t size)
{
for(size_t s = 0; s < size; s++)
{
int a = muis[s];
int testi = 1;
for(size_t n = 0; n < size; n++){
if(s != n) {
if(a == muis[n]){
testi = 0;
break;
}
}
}
if(testi) {
printf("%d \n", a);
}
testi = 1;
}
printf("\n");
}
This approach uses some memory to keep track of which elements are duplicates. The memory cost is higher, but the processor time cost is lower. These differences will become significant at higher values of size.
char* duplicate = calloc(size, 1); // values are initialized to zero
for (unsigned i = 0; i < size; ++i)
{
if(!duplicate[i]) // skip any value that's known to be a duplicate
{
for (unsigned j = i + 1; j < size; ++j) // only look at following values
{
if (array[i] == array[j])
{
duplicate[i] = 1;
duplicate[j] = 1; // all duplicates will be marked
}
}
if (!duplicate[i])
{
printf("%d\n", array[i]);
}
}
}
What you can do is you can initialize a hashmap that will help you store the unique elements. Once you start iterating the array you check for that element in the hashmap. If it is not present in the hashmap add it to the hashmap. If it is already present keep iterating.
This way you would not have to iterate the loop twice. Your time complexity of the algorithm will be O(n).
unordered_map < int, int > map;
for (int i = 0; i < size; i++) {
// Check if present in the hashmap
if (map.find(arr[i]) == map.end()) {
// Insert the element in the hash map
map[arr[i]] = arr[i];
cout << arr[i] << " ";
}
}
Can someone help me to figure out why my code is unable to accurately find the duplicate of elements?
#include <stdio.h>
int main() {
int array[10];
int count = 0;
printf("Enter a maximum of 10 values to store in an array: ");
for (int i = 0; i < 10; i++) {
scanf_s("%d", &array[i]);
}
for (int i = 0; i < 10; i++) {
for (int j = i + 1; j < 10; j++) {
if (array[i] == array[j]) {
count++;
break;
}
}
}
printf("The duplicates are : %d ", count);
}
I'm a beginner at this language so any advice and suggestions to help me solve this exercise will be much appreciated.
First of all the first loop runs 10 times even if the user enters less numbers. You can fix that by doing:
for (int i = 0; scanf_s("%d", &array[i]) == 1 && i < 10; i++);
Then the logic of the other two loops is wrong. I initially got wrong what you meant. I thought you wanted to know how many times a number is duplicated. So I wrote the wrong program and then modified it for your purposes. Here is your program:
#include <stdio.h>
int main() {
int n[10];
int dupes[5], d = 0;
int flag = 1, omg;
for ( omg = 0; scanf("%d", &n[omg]) == 1 && omg < 10; omg++);
for (int i = 0; i < omg; i++) {
for (int j = i+1; j < 10; j++) {
if( n[i] == n[j] ) {
if( d > 0 ) {
for(int k = 0; k < d; k++) {
if( n[i] == dupes[k] ) {
flag = 0;
break;
}
}
}
if( flag ) {
dupes[d] = n[i];
++d;
break;
}
else {
flag = 1;
break;
}
} // end outer if
}
}
printf("There are %d numbers that have at least one dupe\n", d);
return 0;
}
I named a variable omg out of desperation, writing this program was a nightmare. (Because it came from the ashes of a previous program)
Your code correctly determines the number of duplicate entries in the array.
If instead you want to determine the number of duplicated values, you must modify the algorithm:
#include <stdio.h>
int main() {
int array[10] = { 0 };
int count = 0;
printf("Enter a maximum of 10 values to store in an array: ");
for (int i = 0; i < 10; i++) {
scanf_s("%d", &array[i]);
}
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (array[i] == array[j]) {
if (i < j)
count++;
if (i != j)
break;
}
}
}
printf("There are %d duplicate values\n", count);
return 0;
}
I use a structure 'Number' which contains the number and its duplicate, then I fill the array and I put it in ascending order then I calculate the number of duplicate of each number and I fill in the strecture like this :
my code:
#include <stdio.h>
#define size 10
typedef struct Number
{
int number;
int duplicate;
}Number;
int main()
{
int array[size];
Number array2[size];
int count = 0;
printf("Enter a maximum of 10 values to store in an array: ");
for (int i = 0; i < size; i++)
{
scanf("%d", &array[i]);
}
int temp=size;
int temppppp=0;
for(int i=0;i<size;i++)
{
for(int j=i+1;j<size;j++)
{
if(array[i]>array[j])
{
temppppp=array[i];
array[i]=array[j];
array[j]=temppppp;
}
}
}
printf("\n\n");
for (int i = 0; i < size; i++)
{
printf("[%d]",array[i]);
}
printf("\n\n");
int i=0;
int j=0;
while(i<size)
{count=1;
while(i<(size-1)&&array[i]==array[i+1])
{
count++;
i++;
}
if(count>=2)
{
array2[j].number=array[i-1];
array2[j].duplicate=count;
j++;
}
i++;
}
int p=0;
while(p<j)
{
printf("\n[%d] has duplicated %d times !\n",array2[p].number,array2[p].duplicate);
p=p+1;
}
printf("\n\n");
printf("\nThere are %d duplicate values\n", j);
}
How to make for loop to while loop
Is it possible to change for loop in array to while loop?
change
for(i = 0; i < 12; ++i) //for loop for selecting only positive
{
if(a[i] > 0) printf("%d ", a[i]); //display positive only using if statement
}
to
int i = 0, a[12];
while(i < 12)
{
if(a[i] > 0)
{
printf("%d ", a[i]);
}
i++;
}
but it doesn't display the positive same as the negative if changed..
int i;
for(i=0; i < count; i++)
{
// Work item here
}
is functionally equivalent to
int i=0;
while (i < count)
{
// Work item here
i++;
}
Today I got a irritating question in a coding contest. I creamed through the first round but in the second round the following question got me in a trap.
Question: Input N = 4
Output:
1
0 1
1 0 1
0 1 0 1
I tried many things but every time I failed.
Apart from this stupid wrong solution I tried many fancy stuff and failed in the end.
What part of my C knowledge is weak?
If you were given this question how would you solve it?
Seems pretty trivial to me:
int main(int argc, char *argv[])
{
int n = strtol(argv[1], NULL, 10);
for (int i = 0; i < n; i++) {
for (int j = 0; j <= i; j++) {
printf("%d ", i % 2 ? j % 2 : 1 - j % 2);
}
printf("\n");
}
return 0;
}
You panicked. Since the values you print depend on the row and column, use both. And name your variables better.
void printBinaryTriangle(const unsigned int rows) {
for(int row = 1; row <= rows; ++row) {
for (int column = 0; column < row; ++column) {
printf("%d ", (row + column) % 2);
}
putchar('\n');
}
}
This did the job for me.
int i,j,flag;
int num=4;
flag=1;
for(i=0;i<num;i++)
{
for(j=0;j<i+1;j++)
{
printf("%d",(j+flag)%2);
}
if(flag)
{
flag=0;
}
else
{
flag=1;
}
printf("\n");
}
Is this what you want?
for(int k = 0; k < N; k++){
for(int i = 0; i < k+1; i++){
if(i % 2 == k % 2)
printf("1 ");
else
printf("0 ");
};
printf("\n");
}
num = 12
new_string = ''
new_list = ''
while num:
for i in range(1,num+1):
strings_of_ones = new_string+ '1' * i
new_list = list(strings_of_ones)
for j in range(1,len(new_list),2):
new_list[j]='0'
print(' '.join(new_list[::-1]))
num -= 1
Answer in Java:-
public class practice {
public static void main(String args[]) {
int n=4;
for (int i=1 ;i<=n ; i++) {
for (int j=1 ;j<=i ;j++) {
int a=i+j;
if (a % 2==0) {
System.out.print("1"+" ");
} else {
System.out.print("0"+" ");
}
}
System.out.println();
}
}
}
I am trying to write a C code that will print a pyramid structure on screen, something like this.
The corresponding code I've written is something like this.
#include <stdio.h>
#include <stdlib.h>
void printArrayFunc(char arr[9][5]) {
int i, j;
printf("=========================================\nprinting the values\n");
for (i = 0; i < 5; i++) {
for (j = 0; j < 9; j++) {
//printf("arr[%d][%d] = %d\n", i,j, arr[i][j]);
if (arr[i][j] == 1)
printf("*");
else
printf(" ");
}
printf("\n");
}
}
int main() {
int i, j;
char arr[9][5] = {
0
};
printf("============================\nfilling the values\n");
for (i = 0; i < 5; i++) {
for (j = 4 - i; j <= 4 + i; j++) {
arr[i][j] = 1;
// printf("arr[%d][%d]= %d\n",i,j,arr[i][j]);
}
//printf("\n");
}
printArrayFunc(arr);
return 0;
}
It is giving an output like
I know I'm doing some silly mistake but at this moment, I'm not able to find what is going wrong. Let me hear your comments on this.
In the function argument:
char arr[9][5]
In the loop:
for (i = 0; i<5; i++) {
for (j = 0; j<9;j++) {
if (arr[i][j] == 1)
You flipped the position of i and j. i should go from 0 to 9, j from 0 to 5.
if (arr[i][j] == 1)
printf("*");
else
printf(" ");
This statement is giving the garbage value in this statement if if condition is true then it print else statement and when else comes true it prints the garbage value.