I am a beginner in C and I am doing problems in hackerrank and I came across a problem called "plus minus" where the expected output in the problem is not obtained.
Where I got the confusion in floating datatypes in calculation. Here is the problem. We have to give array of numbers contains positive,negative and zero values and input and output should be as follows.
6
-4 3 -9 0 4 1
Sample Output
0.500000
0.333333
0.166667
There are 3 positive numbers, 2 negative numbers, and 1 zero in the array.
The proportions of occurrence are positive: 3/6=0.50000, negative:2/6 =0.33333 and zeros:1/6=0.166667 ..
Here is my code that not getting expected output.
int main()
{
int n,i,p,q,r,a[50],pos,res,zer,neg;
double posres,negres,zerres;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
if(a[i]>0)
++p;
else if(a[i]<0)
++q;
else if(a[i]=0)
++r;
}
posres=p/n;
negres=q/n;
zerres=r/n;
printf("%lf\n%lf\n%lf\n",posres,negres,zerres);
return 0;
}
But my output is:
699317.000000
162833139.000000
0.000000
Where did I go wrong?
Replace a[i]=0, which is assigning value 0 to a[i], with a[i]==0, which is a comparison. And initialize p, q, and r to 0 before the loop. Otherwise you'll get undefined behaviour, e.g. the garbage you see.
int main()
{
int n = 0;
int i = 0;
double zero_count = 0;
double pos_count = 0;
double neg_count = 0;
double num = 0;
scanf("%d", &n);
for(i = 0; i < n; i++) {
scanf("%d", &num);
if(num > 0) {
++pos_count;
} else if(num < 0) {
++neg_count;
} else if(num == 0) {
++zero_count;
}
}
printf("%lf\n%lf\n%lf\n",(pos_count/n), (neg_count/n), (zero_count/n));
return 0;
}
Even if there is only one line of code inside a loop, use braces to wrap the loop. You will save yourself a lot of headache when dealing with massive pieces of code.
Insert spaces between variables and syntactical borders. Compilers will understand what you have written, humans find it hard to read.
Related
I am learning C on my own with a book and I cannot for the life of me figure out how to solve this exercise. I'm obviously looking at it in the wrong way or something. Here is an explanation below.
Listed below are some functions and the main function at the bottom. This program is compiled to generate a certain number of random numbers and determine the min and the max of the random numbers. If you copy and paste this code, you will see how it works. Anyways, an exercise asks me to go to the function "prn_random_numbers()" and change the for loop from "for (i = 1; i < k; ++i)" to for (i = 2; i <= k; ++i). This causes the first line format to print incorrectly. The exercise is to further modify the program in the body of the for loop to get the output to be formatted correctly.
To sum it up, the "prn_random_numbers()" function is written to print out 5 random numbers before moving to the next line. Hence the" i % 5" if statement. Now, for some reason, when you make the slight adjustment to the for loop, as the exercise asks above, it causes the first line to only print 4 numbers before moving to the next line. I have tried a number of things, including trying to force it to print the 5th number, but it only duplicated one of the random numbers. I even tried "i % 4" to see if it would print 4 numbers for each row, but it only prints 3 numbers for the first row instead of 4! So it always prints one less number on the first line than it is supposed to. I have n clue why it is doing that and the book does not give an exercise. Do you have any idea?
Bear with me if you think this is a stupid question. I am just learning on my own and I want to make sure I have a good foundation and understand everything as I learn it, before moving forward. I appreciate any help or advice!
prn_random_numbers(k) /* print k random numbers */
int k;
{
int i, r, smallest, biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for (i = 1; i < k; ++i)
{
if (i % 5 == 0)
printf("\n");
r = rand();
smallest = min(r, smallest);
biggest = max(r, biggest);
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
int main()
{
int n;
printf("Some random numbers are to be printed.\n");
printf("How many would you like to see? ");
scanf("%d", &n);
while (n < 1)
{
printf("ERROR! Please enter a positive integer.\n");
printf("How many would you like to see? ");
scanf("%d", &n);
}
prn_random_numbers(n);
return (EXIT_SUCCESS);
}
the following proposed code:
properly initializes the random number generator
cleanly compiles
properly checks for and handles errors
performs the desired functionality
avoids having to list instructions twice
follows the axiom: Only one statement per line and (at most) one variable declaration per statement.
does not use undefined functions like: max() and min()
and now the proposed code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void prn_random_numbers(int k)
{
int count = 1;
int r;
int smallest;
int biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for ( int i = 2; i <= k; i++, count++)
{
if (count % 5 == 0)
{
count = 0;
printf("\n");
}
r = rand();
smallest = (r < smallest)? r : smallest;
biggest = (r > biggest)? r : biggest;
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
int main( void )
{
int n;
srand( (unsigned)time( NULL ) );
do
{
printf("Please enter a positive integer, greater than 0.\n");
printf("How many would you like to see? ");
if( scanf("%d", &n) != 1 )
{
fprintf( stderr, "scanf for number of random numbers failed\n" );
exit( EXIT_FAILURE );
}
} while( n < 1 );
prn_random_numbers(n);
// in modern C, if the returned value from `main()` is 0 then no `return 0;` statement needed
}
a typical run, no input problems is:
Please enter a positive integer, greater than 0.
How many would you like to see? 20
98697066 2110217332 1247184349 421403769 1643589269
1440322693 985220171 1915371488 1920726601 1637143133
2070012356 541419813 1708523311 1237437366 1058236022
926434075 1422865093 2113527574 626328197 1618571881
20 random numbers printed.
Minimum: 98697066
Maximum: 2113527574
Try to use a debugger to solve your problem, it's easy to use and really helpfull :)
SOLUTION:
Your i variable don't count the number of numbers because it is initialize at 1 (in the for statement), so you need to declare a new variable to count properly.
If you have still a problem:
void prn_random_numbers(int k)
{
int count = 1;
int i, r, smallest, biggest;
r = smallest = biggest = rand();
printf("\n%12d", r);
for (i = 2; i <= k; i++, count++) {
if (count % 5 == 0) {
count = 0;
printf("\n");
}
r = rand();
smallest = min(r, smallest);
biggest = max(r, biggest);
printf("%12d", r);
}
printf("\n\n%d random numbers printed.\n", k);
printf("Minimum:%12d\nMaximum:%12d\n", smallest, biggest);
}
In this file I am trying to make something that adds all numbers up to a number entered by a user. Such as, 4: 1 + 2 + 3 + 4 = 10. So if they enter 4 it returns 10.
When I run the code I get an error message saying my file has stopped working. Do i have an endless loop?
#include "biglib.h"
int main()
{
puts("Enter any number and it will return all the numbers from 1 to your number added together.");
// Asking them for their number
int num;
scanf("%i", num);
// then I run a loop, if num == 0 then the program should break from the loop and return 0 in the main function if not run the code inside the program.
int i;
while(num != 0)
{
// I define "i" to be one less than that of num then as long as "i" is greater than 0 keep running the loop and subtract one at the end of it.
for(i = num - 1; i > 0; i--)
{
// in here I do the addition.
num = num + i;
}
// finally I print out the answer.
printf("%i\n",num);
continue;
}
return 0;
}
Yes, you have an infinite loop. Also the input is not stored in the num variable.
#include "stdio.h"
int main(void) {
puts("Enter any number and it will return all the numbers from 1 to your number added together.");
int num;
scanf("%i", &num);
int sum = 0;
while(num>0){
sum += num;
num -= 1;
}
printf("%i\n",sum);
return 0;
}
Some lines of your code seem odd to me.
Why do you use a while loop to test the value of num ?
Why do you put a continue statement as last while loop instruction ?
Remarks:
Your code does not work for negative number, is it the expected behaviour?
You are not testing the scanf return value, which could cause trouble.
I am pretty sure that you should check the scanf prototype.
Hope these questions will lead you to improve your code.
Thank you yadras fro informing me that I had the scanf outside of the while loop that was the problem and now it works when I do this.
int main()
{
puts("Enter any number and it will return all the numbers from 1 to your number added together.");
int num;
int i;
while(num != 0){
scanf("%i", &num);
for(i = num - 1; i > 0; i--)
{
num = num + i;
}
printf("%i\n",num);
}
return 0;
}
I have a task to sort an integer array into an even number array and odd number array. Then i have to show what digits has been placed in which. However, in my code not all places of the array are occupied, so in the end i receive random numbers when I want to show what the arrays odds and evens contain. Instead of random numbers I would like to have literally nothing added in their place.
I did the following:
int main()
{
int evens[10], whole[10], odds[10], i;
printf("Enter 10 integer(/whole) numbers\n");
for (i = 0; i < 10; i++)
{
scanf("%d", &whole[i]);
if (whole[i] % 2 == 0)
{
evens[i] = whole[i];
else odds[i] = whole[i];
}
}
printf("Your even numbers are the following:\n");
for (i = 0; i < 10; i++)
{
printf("%d\n", evens[i]);
}
printf("Your odd numbers are the following:\n");
for (i = 0; i < 10; i++)
{
printf("%d\n", odds[i]);
}
return 0;
}
and then I get this output for having entered digits from 1 to 10:
Your even numbers are the following:
-1832565259
2
1985901418
4
4200864
6
4200958
8
74
10
Your odd numbers are the following:
1
4200864
3
6356652
5
1986281400
7
1985964450
9
1985901427
So how do i get an odds/even array without these random digits like 1985964450 in between? Is there a command to add literally nothing instead?
You should have a counter of odds and a counter of evens.
int oddcount = 0;
int evencount = 0;
When you decide that a number is even, you use this counter to know where in the array it should go. For example:
if (whole[i] % 2 == 0) {
evens[evencount] = whole[i];
evencount++;
}
Notice that evencount not only gives you the number of even numbers, but since arrays indices begin at zero, it also tells you what is the position of the next even number.
Then you modify your for loops at the end to use the actual number of even and odd numbers that were typed. You can even check for zero and print a specific message such as No even numbers supplied.
Also, unless you have been specifically asked to keep the input numbers in an array, you don't need whole. You can do just like so:
int input;
for (i=0; i<10; i++)
{
scanf("%d", &input);
if (input %2 == 0)
/* ... */
else
/* ... */
}
As a final remark, you should indent your code. Indentation is simply augmenting the number of spaces before the code (like I did inside the ifs above). It is very important to indent your code because it makes the structure of your code clear. For a more comprehensive discussion on this, read here: Importance of code indentation.
It is better to have two variables that represents indexes, one you use to add odd numbers, other to add even numbers. Then you will have two arrays without redundant data :)
int evensIndex = 0;
int oddsIndex = 0;
for (i=0; i<10; i++)
{
scanf("%d", &whole[i]);
if (whole[i] %2 == 0)
{
evens[evensIndex] = whole[i];
evensIndex++;
}
else
{
odds[oddsIndex] = whole[i];
oddsIndex++;
}
}
int main()
{
int evens[10], temp, odds[10], i;
int oddIndex = 0, evenIndex = 0;
printf("Enter 10 integer(/whole) numbers\n");
for (i=0; i<10; i++)
{
scanf("%d", &temp);
if(temp%2)
odd[oddIndex++]=temp;
else
even[evenIdex++]=temp;
}
printf("Your even numbers are the following:\n");
for (i=0; i<10; i++)
printf("%d\n", evens[i]);
printf("Your odd numbers are the following:\n");
for (i=0; i<10; i++)
printf("%d\n", odds[i]);
return 0;
}
Even though this question has been asked a million times I just haven't found an answer that actually helps my case, or I simply can't see the solution.
I've been given the task to make a program that takes in a whole number and counts how many times each digit appears in it and also not showing the same information twice. Since we're working with arrays currently I had to do it with arrays of course so since my code is messy due to my lack of knowledge in C I'll try to explain my thought process along with giving you the code.
After entering a number, I took each digit by dividing the number by 10 and putting those digits into an array, then (since the array is reversed) I reversed the reverse array to get it to look nicer (even though it isn't required). After that, I have a bunch of disgusting for loops in which I try to loop through the whole array while comparing the first element to all the elements again, so for each element of the array, I compare it to each element of the array again. I also add the checked element to a new array after each check so I can primarily check if the element has been compared before so I don't have to do the whole thing again but that's where my problem is. I've tried a ton of manipulations with continue or goto but I just can't find the solution. So I just used **EDIT: return 0 ** to see if my idea was good in the first place and to me it seems that it is , I just lack the knowledge to go back to the top of the for loop. Help me please?
// With return 0 the program stops completely after trying to check the digit 1 since it's been checked already. I want it to continue checking the other ones but with many versions of putting continue, it just didn't do the job. //
/// Tried to make the code look better. ///
#include <stdio.h>
#define MAX 100
int main()
{
int a[MAX];
int b[MAX];
int c[MAX];
int n;
int i;
int j;
int k;
int counter1;
int counter2;
printf("Enter a whole number: ");
scanf("%i",&n);
while (1)
{
for (i=0,counter1=0;n>10;i++)
{
a[i] = n%10;
n=n/10;
counter1+=1;
if (n<10)
a[counter1] = n;
}
break;
}
printf("\nNumber o elements in the array: %i", counter1);
printf("\nElements of the array a:");
for (i=0;i<=counter1;i++)
{
printf("%i ",a[i]);
}
printf("\nElements of the array b:");
for (i=counter1,j=0;i>=0;i--,j++)
{
b[j] = a[i];
}
for (i=0;i<=counter1;i++)
{
printf("%i ",b[i]);
}
for (i=0;i<=counter1;i++)
{
for(k=0;k<=counter1;k++)
{
if(b[i]==c[k])
{
return 0;
}
}
for(j=0,counter2=0; j<=counter1;j++)
{
if (b[j] == b[i])
{
counter2+=1;
}
}
printf("\nThe number %i appears %i time(s)", b[i], counter2);
c[i]=b[i];
}
}
The task at hand is very straightforward and certainly doesn't need convoluted constructions, let alone goto.
Your idea to place the digits in an array is good, but you increment counter too early. (Remember that arrays in C start with index 0.) So let's fix that:
int n = 1144526; // example number, assumed to be positive
int digits[12]; // array of digits
int ndigit = 0;
while (n) {
digits[ndigit++] = n % 10;
n /= 10;
}
(The ++ after ndigit will increment ndigit after using its value. Using it as array index inside square brackets is very common in C.)
We just want to count the digits, so reversing the array really isn't necessary. Now we want to count all digits. We could do that by counting all digits when we see then for the first time, e.g. in 337223, count all 3s first, then all 7s and then all 2s, but that will get complicated quickly. It's much easier to count all 10 digits:
int i, d;
for (d = 0; d < 10; d++) {
int count = 0;
for (i = 0; i < ndigit; i++) {
if (digit[i] == d) count++;
}
if (count) printf("%d occurs %d times.\n", d, count);
}
The outer loop goes over all ten digits. The inner loop counts all occurrences of d in the digit array. If the count is positive, write it out.
If you think about it, you can do better. The digits can only have values from 0 to 9. We can keep an array of counts for each digit and pass the digit array once, counting the digits as you go:
int count[10] = {0};
for (i = 0; i < ndigit; i++) {
count[digit[i]]++;
}
for (i = 0; i < 10; i++) {
if (count[i]) printf("%d occurs %d times.\n", i, count[i]);
}
(Remember that = {0} sets the first element of count explicitly to zero and the rest of the elements implicitly, so that you start off with an array of ten zeroes.)
If you think about it, you don't even need the array digit; you can count the digits right away:
int count[10] = {0};
while (n) {
count[n % 10]++;
n /= 10;
}
for (i = 0; i < 10; i++) {
if (count[i]) printf("%d occurs %d times.\n", i, count[i]);
}
Lastly, a word of advice: If you find yourself reaching for exceptional tools to rescue complicated code for a simple task, take a step back and try to simplify the problem. I have the impression that you have added more complicated you even you don't really understand instead.
For example, your method to count the digits is very confused. For example, what is the array c for? You read from it before writing sensible values to it. Try to implement a very simple solution, don't try to be clever at first and go for a simple solution. Even if that's not what you as a human would do, remeber that computers are good at carrying out stupid tasks fast.
I think what you need is a "continue" instead of a return 0.
for (i=0;i<=counter1;i++) {
for(k=0;k<=counter1;k++) {
if(b[i]==c[k]) {
continue; /* formerly return 0; */
}
for(j=0,counter2=0; j<=counter1;j++)
if (b[j] == b[i]){
counter2+=1;
}
}
Please try and see if this program can help you.
#include <stdio.h>
int main() {
unsigned n;
int arr[30];
printf("Enter a whole number: ");
scanf("%i", &n);
int f = 0;
while(n)
{
int b = n % 10;
arr[f] = b;
n /= 10;
++f;
}
for(int i=0;i<f;i++){
int count=1;
for(int j=i+1;j<=f-1;j++){
if(arr[i]==arr[j] && arr[i]!='\0'){
count++;
arr[j]='\0';
}
}
if(arr[i]!='\0'){
printf("%d is %d times.\n",arr[i],count);
}
}
}
Test
Enter a whole number: 12234445
5 is 1 times.
4 is 3 times.
3 is 1 times.
2 is 2 times.
1 is 1 times.
Here is another offering that uses only one loop to analyse the input. I made other changes which are commented.
#include <stdio.h>
int main(void)
{
int count[10] = { 0 };
int n;
int digit;
int elems = 0;
int diff = 0;
printf("Enter a whole number: ");
if(scanf("%d", &n) != 1 || n < 0) { // used %d, %i can accept octal input
puts("Please enter a positive number"); // always check result of scanf
return 1;
}
do {
elems++; // number of digits entered
digit = n % 10;
if(count[digit] == 0) { // number of different digits
diff++;
}
count[digit]++; // count occurrence of each
n /= 10;
} while(n); // do-while ensures a lone 0 works
printf("Number of digits entered: %d\n", elems);
printf("Number of different digits: %d\n", diff);
printf("Occurrence:\n");
for(n = 0; n < 10; n++) {
if(count[n]) {
printf(" %d of %d\n", count[n], n);
}
}
return 0;
}
Program session:
Enter a whole number: 82773712
Number of digits entered: 8
Number of different digits: 5
Occurrence:
1 of 1
2 of 2
1 of 3
3 of 7
1 of 8
it's me again. I deleted my previous question because it was very poorly asked and I didn't even include any code (i'm new at this site, and new at C). So I need to write a program that prints out the digits smaller than 5 out of a given number, and the number of the digits.
For example: 5427891 should be 421 - 3
The assignment also states that i need to print the numbers smaller than 5 in a recursive function, using void.
This is what I've written so far
#include<stdio.h>
void countNum(int n){
//no idea how to start here
}
int main()
{
int num, count = 0;
scanf("%d", &num);
while(num != 0){
num /= 10;
++count;
}
printf(" - %d\n", count);
}
I've written the main function that counts the number of digits, the idea is that i'll assign (not sure i'm using the right word here) the num integer to CountNum to count the number of digits in the result. However, this is where I got stuck. I don't know how to extract and print the digits <5 in my void function. Any tips?
Edit:
I've tried a different method (without using void and starting all over again), but now i get the digits I need, except in reverse. For example, instead of printing out 1324 i get 4231.
Here is the code
#include <stdio.h>
int rec(int num){
if (num==0) {
return 0;
}
int dg=0;
if(num%10<5){
printf("%d", num%10);
dg++;
}
return rec(num/10);
}
int main(){
int n;
scanf("%d", &n);
int i,a;
for(i=0;i<n;i++)
{
scanf("%d", &a);
rec(a);
printf(" \n");
}
return 0;
}
Why is this happening and how should I fix it?
There is nothing in your question that specifies the digits being input are part of an actual int. Rather, its just a sequence of chars that happen to (hopefully) be somewhere in { 0..9 } and in so being, represent some non-bounded number.
That said, you can send as many digit-chars as you like to the following, be it one or a million, makes no difference. As soon as a non-digit or EOF from stdin is encountered, the algorithm will unwind and accumulate the total you seek.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int countDigitsLessThanFive()
{
int c = fgetc(stdin);
if (c == EOF || !isdigit((unsigned char)c))
return 0;
if (c < '5')
{
fputc(c, stdout);
return 1 + countDigitsLessThanFive();
}
return countDigitsLessThanFive();
}
int main()
{
printf(" - %d\n", countDigitsLessThanFive());
return EXIT_SUCCESS;
}
Sample Input/Output
1239872462934800192830823978492387428012983
1232423400123023423420123 - 25
12398724629348001928308239784923874280129831239872462934800192830823978492387428012983
12324234001230234234201231232423400123023423420123 - 50
I somewhat suspect this is not what you're looking for, but I'll leave it here long enough to have you take a peek before dropping it. This algorithm is fairly pointless for a useful demonstration of recursion, to be honest, but at least demonstrates recursion none-the-less.
Modified to print values from most significant to least.
Use the remainder operator %.
"The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined" C11dr ยง6.5.5
On each recursion, find the least significant digit and test it. then divide the number by 10 and recurse if needed. Print this value, if any, after the recursive call.
static int PrintSmallDigit_r(int num) {
int count = 0;
int digit = abs(num % 10);
num /= 10;
if (num) {
count = PrintSmallDigit_r(num);
}
if (digit < 5) {
count++;
putc(digit + '0', stdout);
}
return count;
}
void PrintSmallDigits(int num) {
printf(" - %d\n", PrintSmallDigit_r(num));
}
int main(void) {
PrintSmallDigits(5427891);
PrintSmallDigits(-5427891);
PrintSmallDigits(0);
return 0;
}
Output
421 - 3
421 - 3
0 - 1
Notes:
This approach works for 0 and negative numbers.
First of all, what you wrote is not a recursion. The idea is that the function will call itself with the less number of digits every time until it'll check them all.
Here is a snippet which might help you to understand the idea:
int countNum(int val)
{
if(!val) return 0;
return countNum(val/10) + ((val % 10) < 5);
}
void countNum(int n, int *c){
if(n != 0){
int num = n % 10;
countNum(n / 10, c);
if(num < 5){
printf("%d", num);
++*c;
}
}
}
int main(){
int num, count = 0;
scanf("%d", &num);
countNum(num, &count);
printf(" - %d\n", count);
return 0;
}
for UPDATE
int rec(int num){
if (num==0) {
return 0;
}
int dg;
dg = rec(num/10);//The order in which you call.
if(num%10<5){
printf("%d", num%10);
dg++;
}
return dg;
}
int main(){
int n;
scanf("%d", &n);
int i,a;
for(i=0;i<n;i++){
scanf("%d", &a);
printf(" - %d\n", rec(a));
}
return 0;
}