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Can someone tell what wrong with my code?
This code is attempting to deduce a palindrome for numbers from 100-999
#include <stdio.h>
int main ()
{
int temp, n, reverse = 0, rem;
n = 100;
while ( n <= 999)
{
temp = n;
while(temp!=0)
{
rem = temp%10;
reverse = reverse*10+rem;
temp/=10;
}
if ( reverse == n )
{
printf(" %d \t", n);
}
n = n + 1;
}
return 0;
}
Thanks.
You should be setting reverse back to 0 for each new n
while ( n <= 999)
{
reverse = 0; # <== otherwise you are using old value in reverse
temp = n;
while(temp!=0)
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This might be simple but I'm new to recursion in c. I want to find the sum of odd integers based on user's input. For example if user inputs 3, function returns 9 (1 + 3 + 5 = 9)
int recursiveSumNOdd(int n)
{
int start = -2; //later start = start+2, so it starts at 0
int n1 = n*2; //total number of digits with rec
int num = 0;
int sum=0;
int count=0;
if(start>=n1)
{
return 0;
}
else
{
start = start+2;
count++;
sum = sum +recursiveSumNOdd(start);
}
return sum;
}
Explanations in comment:
int recursiveSumNOdd(int n) {
if (n == 1 || n == 0)// first "if" in a recursive is its stop condition
return n;
return 2 * n - 1 + recursiveSumNOdd(n-1); // formula for 2->3, 3->5 etc
}
int main(void) {
printf("%d\n", recursiveSumNOdd(3));
return 0;
}
NB: You may want to handle integer overflow
NB2: You can have a mathematics formula to return instantly the result, it is way better, but I guess it was to understand better recursion?
return n * n; // the sum of odd numbers is the square of user's input
You are over-complicating things.
You cannot have the sum of negative elements.
int sum_n_odd(unsigned n)
{
What is the sum of 0 (zero) elements?
if (n == 0) return 0;
If you knew the sum of n - 1 numbers, what is the sum of n numbers?
return sum_n_odd(n - 1) + something; // something is easy to figure out
}
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I was given code and I don't understand why the function returns -1. I feel that it is a typo. The code is given below:
int equilibrium(int array[], int size)
{
int sum = 0;
int left_sum = 0;
int i;
for (i = 0; i < size; i++) {
sum += array[i];
}
for (i = 0; i < size; i++) {
if (array[i] == sum - 2 * left_sum) {
return i;
}
left_sum += array[i];
}
return -1;
}
It looks like it uses -1 to mean "not found", it looks like a search function trying to find an index i where the condition in the innermost if is true.
equilibrium returns the equilibrium point of an array as an index in the array. When the array doesn't have an equilibrium point, the function returns -1 instead. It is used as followed:
int equilibrium_idx = equilibrium(somearrayvalue, somesize);
if( equilibrium_idx == -1 )
printf("It isn't balanced\n");
else
printf("The equilibrium index is %d\n", equilibrium_idx);
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Using C Language:
How can I generate random numbers in the [pi, 2pi] range?
How can I generate random numbers in the [-1, 0] range?
This is a pseudo random integer generator, but has only been tested for positive numbers, you would have to modify it to use with negatives:
int randomGenerator(int min, int max)
{
int random=0, trying=0;
trying = 1;
srand(clock());
while(trying)
{
random = (rand()/32767.0)*(max+1);
(random >= min) ? (trying = 0) : (trying = 1);
}
return random;
}
EDIT (last method did not produce randoms due to srand() not being updated enough)
For a range spanning positive and negative numbers you could modify it like this: (but ratio of pos to neg would be same)
int randomGenerator(int min, int max)
{
int random=0, trying=0;
int i=0;
trying = 1;
srand(clock());
while(trying)
{
random = (rand()/32767.0)*(max+1);
(random >= min) ? (trying = 0) : (trying = 1);
}
return (i++%2==0)?(random):(-1*random);
}
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Assume the character a,b,c,d,e represent the number 1 to 9, and they cannot be equal to
each other.
Question:
How many equals that can meet (ab * cde = adb * ce).
Example:
36 * 495 = 396 * 45.
Here is my code,and the result is right.However,i think my code is too awkward,especially in (if(a!=b&&a!=c&&a!=d&&a!=e&&b!=c&&b!=d&&b!=e&&c!=d&&c!=e&&d!=e&&c*d*e!=0))
I would appreciate it if someone could give me a better solution.
#include<stdio.h>
main(){
int a,b,c,d,e,m,n,i=0;
long f1,f2,f3,f4;
for(m=11;m<=99;m++){
a=m/10;
b=m%10;
if(a!=b&&a*b!=0)
{
for(n=101;n<=999;n++)
{
c=n/100;
d=n%100/10;
e=n%10;
if(a!=b&&a!=c&&a!=d&&a!=e&&b!=c&&b!=d&&b!=e&&c!=d&&c!=e&&d!=e&&c*d*e!=0)
{
f1=a*10+b;
f2=c*100+d*10+e;
f3=a*100+d*10+b;
f4=c*10+e;
if(f1*f2==f3*f4) i++;
printf("\n%d%d*%d%d%d*=%d%d%d*%d%d\n",a,b,c,d,e,a,d,b,c,e);
}
}
}
}
printf("%d\n",i);
return 0;
}
If you can, instead of
int a,b,c,d,e;
Try to use
int numbers[5];
And then to check if your numbers are all different, you can use for loops
doubleOccurence = FALSE; /* where FALSE = 0 */
for (i=0; i < 4; i++) {
for (j=i+1; j < 5; j++) {
doubleOccurence = doubleOccurence || (numbers[i] == numbers[j]);
}
}
It looks a bit clearer to me.
Unfortunately you can't really iterate through a list of variables you are better off with an array of numbers like Julien mentions in his answer.
int nums[5];
replace a with nums[0], b with nums[1], etc....
But then I would go one step further to tidying up your code and call a function that takes in the array to check uniqueness:
if(listIsUnique(nums, 5)) // yes hardcoded the 5, but that can be sorted
{
...
}
And then:
bool listIsUnique(int* nums, int len)
{
for (int i = 0; i < len; i++)
for (int j = i + 1; j < len; j++)
if (nums[i] == nums[j])
return false; // return false as soon as you find a match - slightly faster :)
return true; // if we get here its a unique list :)
}
Note: code is untested, there may be mistakes :o
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I'm having trouble with arrays. I copied this code from a book:
#include <stdio.h>
#include <stdbool.h>
int main (void)
{
int p, i, primes[50], primeIndex = 2;
bool isPrime;
primes[0] = 2;
primes[1] = 3;
for (p = 5; p <= 50; p = p + 2) {
isPrime = true;
for (i = 1; isPrime && p / primes[i] >= primes[i]; ++i)
if (p % primes[i] == 0)
isPrime = false;
if (isPrime == true) {
primes[primeIndex] = p;
++primeIndex;
}
}
for (i = 0; i < primeIndex; ++i)
printf ("%i ", primes[i]);
printf ("\n");
return 0;
}
In particular, I'm having trouble understanding the difference between the primeIndex and the i variables. The primeIndex refers to the array number and i refers to the number placed into the array. Right?
primeIndex is the place where the next found prime is written in the prime array, and also the number of primes known so far. i is the index of the prime used for trial division. For each candidate, i loops from 1 (we don't need to try out primes[0] = 2 because only odd numbers are checked) to the index of the first prime larger than the square root of the candidate.