I am attempting to write a C program that can calculate the factorial of a number inputted by a user. I have the program below, and it seems to work fine without any errors. However, everything I input a number like 5, I get a strange answer like -1899959296.
I cant seem to find where the error is, let alone how the answer is negative. Any help would be appreciated, thank you very much.
int main()
{
int inputnumber;
int n;
printf("\nThis program is a factorial calculator. Input a number below to finds its factorial: ");
scanf("%i", &inputnumber);
//A 'for' loop repeats the same contained statements until a condition is met.
//A general form of a 'for' statements is shown below:
// for( initial values; loop condition; loop expression)
for (n = 1; n <= inputnumber; n++)
{
inputnumber = inputnumber * n;
}
printf("\nThe answer is %i", inputnumber);
return 0;
}
I attempted to make a factorial program, but all inputted numbers give very wrong answers.
The problem is this part of your code:
for (n = 1; n <= inputnumber; n++)
{
inputnumber = inputnumber * n;
}
The code inside the loop increases inputnumber when n > 1. This causes the loop to run much longer than intended because n <= inputnumber (your loop conditional) is always satisfied until there is an integer overflow, i.e., when inputnumber < 0. Hence, the loop only stops when inputnumber becomes negative.
To fix this, you should store your factorial result as a separate variable, e.g.:
int result = 1;
Then update your loop to be result *= n; (and obviously change your print statement accordingly).
If you changed your conditional portion of your for loop to restrict to only < rather than <= it should produce the right answers.
For example i assume the input: '2' gives you '4' as a result:
input = 2 * 1
input = 2 * 2
end
return input
or by using a third variable you can use your current logic:
#include <stdio.h>
int main() {
int number, i, factorial = 1;
printf("Enter an integer: ");
scanf("%d", &number);
for (i = 1; i <= number; i++) {
factorial *= i;
}
printf("Factorial of %d = %d\n", number, factorial);
return 0;
}
Related
Something is wrong. I'm trying to make a cade which can count number count of any natural number. Like number count of 2 is 1, 30 is 2, 456 is 3. My code is running for 1 digit numbers but not for two digit numbers.
#include<stdio.h>
void main(void)
{
int num,count,check;
float div;
printf("Enter a natural number\n");
scanf("%d", &num);
while (num<=0)
{
printf("Error\n");
printf("Enter a number\n");
scanf("%d", &num);
}
while(num>1)
{
count=1;
check=10;
div=num/check;
if(div<=1)
{
printf("Number count is\n%d", count);
break;
}
check = check*10;
count = count+1;
}
}
The problem with your solution is that after check and count are modified at the end of the loop, they are re-declared to 1 and 10 respectively at the beginning of the loop at every passage.
You need to move the declaration just before the while loop.
Also div doesn't need to be a float given that the decimal part of this number is irrelevant.
You could also use less variables by replacing check by 10 and
using num directly instead of temporarily storing results in div.
I think this might be a simpler solution
#include <stdio.h>
int main(){
int num = 0, digits = 0;
while (num <= 0)
{
printf("Enter a natural number\n");
scanf("%d", &num);
num == 0 ? printf("Error\n") : 0;
}
for( ; num > 0; digits++)
num /= 10;
printf("number of digits: %d\n", digits);
}
As num is continuously divided by 10, the decimal of the result gets truncated since num is an int while digits steadily increases.
It is time to learn to use a debugger. Using it would have immediately shown the major problem in your code: you reset the value of count and check inside the loop. So if you enter a number greater or equal to 10, you enter an infinite loop because you will consistently divide that number by 10 and find that the result is >= 1!
There is another less important problem: you use if(div<=1) when it should be if(div<1). Because 10/10 is 1 and has 2 digits...
After those fixes you should have:
...
check = 10;
count = 1;
while (num > 1)
{
div = num / check;
if (div < 1)
{
printf("Number count is\n%d", count);
break;
}
check = check * 10;
count = count + 1;
}
return 0; // main shall return an int value to its environment...
}
Which correctly gives the number of decimal digit on positive integers. But as you were said in comments, you should always test the return value of scanf (what is the user inadvertently types a t instead of 5 for example?).
That being said, this answer intends to show you what the problems were, but Keyne's solution is better...
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 am really stuck with this been trying to solve it for quite a time now.
i have to write a program where i should input 5 numbers between 1 to 10 and then calculate the average, USING ONLY WHILE LOOP, but it does not have to exit when the number does not meet the requirement. then, i have to write a variation of the same code but this time you can enter all the numbers you want, and when 0 is entered it has to calculate the average and exit
this is where i have gotten so far
#include <stdio.h>
int main(void)
{
int n, i = 1;
float add;
float avg;
do
{
printf("enter the number %d:\n", i++);
scanf("%d", &n);
add = add + n;
} while(n > 0 && n < 11);
avg= (add / 5);
printf("%.1f", avg);
return 0;
}
it will keep asking for numbers after 5 have been entered. and the average is not right anyways
First, you're using nas your while condition variable, but also as the variable to scan the input. If I start your program by scanning 20, for example, your while loop will exit on the first interaction. Use your i variable instead and also increment it every time your loop executes.
do{
...
}while(i <= 5);
Second, if you want only numbers between 1 and 10, then you should write a condition for it. For example:
printf("enter the number %d:\n", i); //do not increment it here!
scanf("%d",&n); //assuming "n" as your variable to scan
if(n > 0 && n < 11){
add += n;
i++; //increment it here instead!
}
Third, initialize your variables in order to not get thrash values
float add = 0;
float avg = 0;
int i = 1;
Finally, assign your result (not mandatory, but since you're using it I'll keep it):
avg = add/5.0f
and display:
printf("%.1f", avg);
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
I am solving problem in a online judge and I faced a problem. I think my code is correct but unfortunately the judge says wrong answer. Where did I made mistake?
Habib has learned so much about programming in the last few days! Today he faces a new challenge, handling input with test cases! For this, he must solve a set of inputs and for each input h must generate an output. And what is better than to practice this other than calculating factorials?! A factorial of an integer N is calculated by multiplying all the integers from 1 to N. For example, 4! (4 factorial) is calculated as- 1x2x3x4=24. In this problem, Habib is required to solve a set of inputs for a defined number of test cases. For example, if testcase = 3, then he must take 3 set of inputs and generate 3 sets of desired outputs, one output for one input. Help him solve the problem.
Input
The input starts with an integer Test (0 < Test < 100) which denoted number of inputs or testcases to be solved. For each testcase, input an integer n (0 <= n <= 10).
Output
For each testcase generate output in a single line in the format: Case x: y, where x is the testcase number and y is the answer for calculating n!.
#include <stdio.h>
int main()
{
int i,Test,n=0,x,j,y,s=1;
scanf("%d",&Test);
for(i=1;i<=Test;i++)
{
scanf("%d",&n);
for(j=1;j<=n;j++)
{
s=s*j;
}
printf("Case %d: %d\n",i,s);
}
return 0;
}
You forgot to reset s for each test cases.
Try adding s=1; after scanf("%d",&n);.
Input The input starts with an integer Test (0 < Test < 100) which denoted number of inputs or testcases to be solved. For each testcase, input an integer n (0 <= n <= 10).
Your code does not reflect this. You should definitely check if the inputs are valid or not, a way could be:
if(Test < 100 && Test > 0){
//do something
}else{
printf("Invalid input!");
}
Furthermore you should consider staying with either capital variable names or only lowercase, example: int test,n or int Test, N. Mixing both makes bigger projects a guessing game when you have to edit it after a few days.
Now to your Task:
Basically your task is to find n! and do this X-times.
As stated by #MikeCat you never reset your s, which is the starting point for your factorial.
#include <stdio.h>
int main(){
int i,test,n=0,x,j,y,s=1;
scanf("%d",&Test);
if(!(test <100 && test > 0)){
printf("Invalid input!");
}
for(i=1;i<=test;i++)
{
scanf("%d",&n);
s = 1;
for(j=1;j<=n;j++)
{
s=s*j;
}
printf("Case %d: %d\n",i,s);
}
return 0;
}
This should do the trick.
Change your code to
#include <stdio.h>
int main()
{
int i,Test,n=0,x,j,y;
scanf("%d",&Test);
for(i=1;i<=Test;i++)
{
int s=1;//observe this
scanf("%d",&n);
for(j=1;j<=n;j++)
{
s=s*j;
}
printf("Case %d: %d\n",i,s);
}
return 0;
}