Doing an exercise to take 5 integers from user and add up only the odd numbers. Everything adds up correctly until the last number messes everything up for some reason:
Code and Test
#include <stdio.h>
int main() {
int userNum[5];
int i;
int sum = 0;
for (i = 1; i <= 5; ++i) {
printf("Please enter number %d:\n", i);
scanf("%d", &userNum[i]);
if (userNum[i] % 2 > 0) {
sum = sum + userNum[i];
}
}
printf("The sum of all odd integers is: %d", sum);
return 0;
}
You are reading the 5 numbers into the array elements userNum[1] through userNum[5], but since array indices start at 0 in C, userNum[5] does not exist and the program has undefined behavior when it attempts to store a number beyond the end of the array. Some other variable gets modified and the output is bogus. Undefined behavior could actually have much worse consequences, such as a bogus candidate winning an election with fewer votes than his opponent :)
Here is a corrected version:
#include <stdio.h>
int main() {
int userNum[5];
int i;
int sum = 0;
for (i = 0; i < 5; ++i) {
printf("Please enter number %d:\n", i + 1);
if (scanf("%d", &userNum[i]) != 1) {
printf("invalid input\n");
return 1;
}
if (userNum[i] % 2 > 0) {
sum = sum + userNum[i];
}
}
printf("The sum of all odd integers is: %d\n", sum);
return 0;
}
Related
I have return the code to find a factorial and to display trailing zeros at the end of the factorial, but the output is wrong... could you please help me to find the mistake?
#include <stdio.h>
int main() {
int m = 1, i, N, count = 0;
scanf("%d", &N);
for (i = 1; i <= N; i++) {
m = m * i;
}
printf("%d", m);
while (m > 0) {
if ((m % 10) == 0) {
count = count + 1;
m = m / 10;
}
break;
}
printf("%d", count);
return 0;
}
Your code only works for very small values of N: up to 9. For slightly larger values, you would need to add an else keyword before the break statement and you would get a correct result for a few more cases.
For larger values, you must compute the power of 5 that divides the factorial. You can do this incrementally by summing the power of 5 that divide each individual number up to and including N.
#include <stdio.h>
int main() {
int N, count;
if (scanf("%d", &N) != 1)
return 1;
/* only consider factors that are multiples of 5 */
count = 0;
for (int i = 5; i <= N; i += 5) {
for (int j = i; j % 5 == 0; j /= 5)
count++;
}
printf("%d\n", count);
return 0;
}
An even simpler and faster solution is this: compute the number of multiples of 5 less or equal to N, add the number of multiples of 5*5, etc.
Here is the code:
#include <stdio.h>
int main() {
int N, count;
if (scanf("%d", &N) != 1)
return 1;
count = 0;
for (int i = N; (i /= 5) > 0;) {
count += i;
}
printf("%d\n", count);
return 0;
}
you have two problems
your collapse the two outputs so you see only one of them / you cannot see who is who, just add a separator between them
an else is missing when you count so you count to only up to 1 and the result is wrong from factorial 10
So the minimal changes produce :
int main()
{
int m=1,i,N,count=0;
scanf("%d",&N);
for(i=1;i<=N;i++)
{
m=m*i;
}
printf("%d\n",m); /* <<< added \n */
while(m>0)
{
if((m%10)==0)
{
count=count+1;
m=m/10;
}
else /* <<< added else */
break;
}
printf("%d\n",count); /* <<< added \n */
return 0;
}
after the changes :
pi#raspberrypi:/tmp $ ./a.out
5
120
1
pi#raspberrypi:/tmp $ ./a.out
10
3628800
2
Of course that supposes first you are able to compute the factorial without overflow
I also encourage you to check a value was read by scanf, checking it returns 1
#include <stdio.h>
int main()
{
int n,i,f=1,t,c=0;
printf("Enter number ");
scanf("%d",&n);
t=n;
for(i=1;t>=5;i++)
{
t=n/5;
c=c+t;
n=t;
}
printf("number of zeros are %d",c);
return 0;
}
#include <stdio.h>
int main()
{
int i; //counter for the loop
int n; //integer
int series;
printf("Enter an integer number: ");
scanf("%d" , &n);
for(i = 1; i <= n; i++)
{
if (i % 2 == 0)
(series -= i * i);
else
(series += i * i);
}
printf("The value of the series is: %d\n" , series);
return 0;
}
So the the loop is just a basic for loop, using I as the counter for as long as it is less than or equal to n
the series that I have to replicate adds odd numbers and subtracts even numbers so the if condition tests if the number is even or odd. The program compiles fine but when I enter the integer as 5 the sum of the series should be 15, however my program gives the sum 32779. Any help on fixing my program would be appreciated.
you didn't initialize series, so it's a random value in the beginning of the calculation.
#include <stdio.h>
int main()
{
int i = 0; //counter for the loop
int n = 0; //integer
int series = 0;
printf("Enter an integer number: ");
scanf("%d" , &n);
for(i = 1; i <= n; i++)
{
if (i % 2 == 0)
(series -= i * i);
else
(series += i * i);
}
printf("The value of the series is: %d\n" , series);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int summation(int sum,int Num[],int n) {
if (n < 0) {
return sum;
} else {
printf("%d \n", Num[n]);
sum = Num[n] + summation(sum, Num, n - 1);
}
}
int main () {
int i = 0, j = 0 , k = 0, n = 0;
printf("Enter number of numbers: ");
scanf("%d", &n);
int Num[n];
while (i < n) {
//printf("Enter number %d : ", i);
//scanf("%d", &Num[i]);
Num[i] = 1;
i++;
}
int sum = 0;
sum = summation(sum, Num, n);
printf("The sum is %d \n", sum);
}
The above program runs correctly only with 3
why does it generate only results for 3,3^3, 3^9??
Please help me out here this programs runs only with powers of 3. For now i have calculated 3^1,3^3 and 3^9. For other numbers it returns a wrong value.
You neglect to return a value from the summation function. A good compiler would draw your attention to this fact.
You call summation with n equal to the size of the function, which causes this line:
sum = Num[n] + ...
to read past the end of the array.
These mistakes lead to undefined behavior, which can show odd preferences for powers of three, or anything else.
Well, the problem is the above. To sum it up, it compiles, but I guess my main idea is just wrong. What I'm trying to do with that code is:
I want the person to give us the elements of the array, how many he wants to (with a limit of a 100 elements).
After that, I'm checking what array positions are prime numbers.(ex: position 2,3,5,etc. Not the elements itself).
After that, I'm doing the average of the values in the prime numbers position.
That's it. Any ideas? Keep in mind that I'm on the first period of engineering, so I'm not the best in programming.
The code is below:
#include <stdio.h>
#include <windows.h>
int main(void)
{
int k, i, j, d, v[101], sum, prim, f;
float ave;
i = 0;
while ((i < 101) && (j != 0) )
{
i++;
printf("Set the value of the element %d => ", i);
scanf("%d", &v[i]);
printf("To stop press 0 => ");
scanf("%d", &j);
}
k = 0;
prim = 1;
f = i;
sum = 0;
while (f > 2)
{
if (i % (f - 1) == 0)
{
prim = 0;
}
else
{
k++;
sum = sum + v[f];
}
f = f - 1;
}
med = sum / k;
printf("%d, %d, %d", k, soma, i);
printf("The average is => %f \n", ave);
system("pause");
}
For those wondering, this is what i got after the editing in the correct answer:
int main(void)
{
int v[101];
int n = 0;
int k,j = 0;
int i=0;
int sum = 0;
while( i<100 )
{
i++;
printf ("Set the value of the element %d => ", i);
scanf ("%d", &v[i]);
int x,primo=1;
if (i>1){
for (x=2; x*x<=i; x++) {
if (i % x == 0) primo = 0;
}
if(primo==1)
{
sum = sum+ v[i];
n++;
}
}
printf ("To stop press 0 => ");
scanf ("%d", &j);
if(j == 0)
break;
}
float ave =(sum /n);
printf("%d, %d, %d", n,i,sum);
printf("The average is => %f \n", ave);
system("pause");
}
First lets make a readable method to test if a number is prime; this answer from another SO post gives us a good one:
int IsPrime(int number) {
int i;
for (i=2; i*i<=number; i++) {
if (number % i == 0) return 0;
}
return 1;
}
Second, let's clean your code, and compute a running sum of all the prime numbers encountered so far. Also, we will check the return values of scanf (but we should avoid scanf !)
And third, we add some indentation.
int main(void)
{
int n = 0;
int i = 0;
int j = 0;
int k = 0;
int sum = 0;
while( i<101 )
{
i++;
printf ("Set the value of the element %d => ", i);
if(scanf ("%d", &k) != 1)
continue;
if(is_prime(k))
{
sum += k;
++n;
}
printf ("To stop press 0 => ");
if(scanf ("%d", &j) == 1)
if(j == 0)
break;
}
float ave = sum / (double) n;
printf("The average is => %f \n", ave);
system("pause");
}
Well there are a few things to say. First the easy part: if the max number of integers allowed to read is 100 your variable "v" should be v[100]. This is not a char array, so this array don't need to have an extra element (v[100] will be an array of int that goes from v[0] to v[99]; adjust the loop limit too).
Also, you are checking if the number you have is prime in the variable f, but this var is assigned with the variable i and i is not an element of the array. You want to assign f something like v[i] (for i equal to 0 to the count of numbers read minus one). So you will need 2 loops: the one you are using now for checking if the number is prime, and another one that assigns v[i] to f.
Another thing to say is that you are calling scanf two times for reading, you could just read numbers and store it in a temporary variable. If this number is not zero then you store it in the array and keep reading, else you stop the reading.
By last I strongly recommend you set var names that make sense, use single letters only for the index variables; names like temp, array, max and countnumbers should appear in your code. It will be easier for you and everyone else to read your code, and you will reduce the number of mistakes.
Here's the solution to your problem. Very easy stuff.
/* C program to find average of all prime numbers from the inputted array(you can predefine it if you like.) */
#include <stdio.h>
#include <conio.h>
void main()
{
int ar[100], i, n, j, counter;
float avg = 0, numprime = 0;
printf("Enter the size of the array ");
scanf("%d", &n);
printf("\n Now enter the elements of the array");
for (i = 0; i < n; i++)
{
scanf("%d", &ar[i]);
}
printf(" Array is -");
for (i = 0; i < n; i++)
{
printf("\t %d", ar[i]);
}
printf("\n All the prime numbers in the array are -");
for (i = 0; i < n; i++)
{
counter = 0;
for (j = 2; j < ar[i]; j++)
{
if (ar[i] % j == 0)
{
counter = 1;
break;
}
}
if (counter == 0)
{
printf("\t %d", ar[i]);
numprime += 1;
avg += at[i];
}
}
avg /= numprime;
printf("Average of prime numbers is ℅f", avg);
getch();
}
You just need counter variables like above for all average computations. (Cause we need to know number of prime numbers in the array so we can divide the total of them and thus get average.) Don't worry about typecasting it is being done downwards... This solution works. I've written it myself.
Here is a cut at doing what you wanted. You don't need near the number of variables you originally had. Also, without knowing what you wanted to do with the prime number, I just output when a prime was encountered. Also as previously mentioned, using a function for checking prime really helps:
#include <stdio.h>
// #include <windows.h>
/* see: http://stackoverflow.com/questions/1538644/c-determine-if-a-number-is-prime */
int IsPrime(unsigned int number) {
if (number <= 1) return 0; // zero and one are not prime
unsigned int i;
for (i=2; i*i<=number; i++) {
if (number % i == 0) return 0;
}
return 1;
}
int main(void)
{
int i, v[101], sum, pcnt=0, psum=0;
float ave;
i=0;
printf ("\nEnter array values below, use [ctrl + d] to end input\n\n");
printf ("Set the value of the element %d => ", i);
while((i<101) && scanf ("%d", &v[i]) != EOF ){
sum += v[i];
if (IsPrime (v[i]))
psum += v[i], pcnt++;
i++;
printf ("Set the value of the element %d => ", i);
}
ave=(float)psum/pcnt;
printf("\n\n Number of elements : %d\n",i);
printf(" The sum of the elements: %d\n",sum);
printf(" The number of primes : %d\n",pcnt);
printf(" The average of primes : %f\n\n", ave);
return 0;
}
Sample Output:
Enter array values below, use [ctrl + d] to end input
Set the value of the element 0 => 10
Set the value of the element 1 => 20
Set the value of the element 2 => 30
Set the value of the element 3 => 40
Set the value of the element 4 => 51
Set the value of the element 5 => 11
Set the value of the element 6 => 37
Set the value of the element 7 =>
Number of elements : 7
The sum of the elements: 199
The number of primes : 2
The average of primes : 24.000000
I'm very new to programming and I was asked to find the sum of prime numbers in a given range, using a while loop. If The input is 5, the answer should be 28 (2+3+5+7+11). I tried writing the code but it seems that the logic isn't right.
CODE
#include <stdio.h>
int main()
{
int range,test;
int sum = 2;
int n = 3;
printf("Enter the range.");
scanf("%i",range);
while (range > 0)
{
int i =2;
while(i<n)
{
test = n%i;
if (test==0)
{
goto end;
}
i++;
}
if (test != 0)
{
sum = sum + test;
range--;
}
end:
n++;
}
printf("The sum is %i",sum);
return 0;
}
It would be nice if you could point out my mistake and possibly tell me how to go about from there.
first of all, in the scanf use &range and not range
scanf("%i",&range);
Second this instruction is not correct
sum = sum + test;
it should be
sum = sum + n;
and also the
while (range > 0)
should be changed to
while (range > 1)
Because in your algorithm you have already put the first element of the range in the sum sum = 2 so the while should loop range - 1 times and not range times
That's all
OK, my C is really bad, but try something like the following code. Probably doesn't compile, but if it's a homework or something, you better figure it out yourself:
UPDATE: Made it a while loop as requested.
#include <stdio.h>
int main()
{
int range, test, counter, innerCounter, sum = 1;
int countPrimes = 1;
int [50] primesArray;
primesArray[0] = 1;
printf("Enter the range.");
scanf("%i",range);
counter = 2;
while (counter <= range) {
for (innerCounter = 1; innerCounter < countPrimes; innerCounter++) {
if (counter % primesArray[innerCounter] == 0)
continue;
primesArray[countPrimes + 1] = counter;
countPrimes ++;
sum += counter;
}
counter ++
}
printf("The sum is %i",sum);
return 0;
}
I haven't done C in a while, but I'd make a few functions to simplify your logic:
#include <stdio.h>
#include <math.h>
int is_prime(n) {
int i;
for (i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
return 0;
}
}
return 1;
}
int main() {
int range, i, sum, num_primes = 0;
printf("Enter the range: ");
scanf("%d", &range);
for (i = 2; num_primes < range; i++) {
if (is_prime(i)) {
sum += i;
num_primes++;
}
}
printf("The sum is %d", sum);
return 0;
}
Using goto and shoving all of your code into main() will make your program hard to debug.
Copy - pasted from here.
#include <stdio.h>
int main() {
int i, n, count = 0, value = 2, flag = 1, total = 0;
/* get the input value n from the user */
printf("Enter the value for n:");
scanf("%d", &n);
/* calculate the sum of first n prime nos */
while (count < n) {
for (i = 2; i <= value - 1; i++) {
if (value % i == 0) {
flag = 0;
break;
}
}
if (flag) {
total = total + value;
count++;
}
value++;
flag = 1;
}
/* print the sum of first n prime numbers */
printf("Sum of first %d prime numbers is %d\n", n, total);
return 0;
}
Output:
Enter the value for n:5
Sum of first 5 prime numbers is 28
Try the simplest approach over here. Check C program to find sum of all prime between 1 and n numbers.
CODE
#include <stdio.h>
int main()
{
int i, j, n, isPrime, sum=0;
/*
* Reads a number from user
*/
printf("Find sum of all prime between 1 to : ");
scanf("%d", &n);
/*
* Finds all prime numbers between 1 to n
*/
for(i=2; i<=n; i++)
{
/*
* Checks if the current number i is Prime or not
*/
isPrime = 1;
for(j=2; j<=i/2 ;j++)
{
if(i%j==0)
{
isPrime = 0;
break;
}
}
/*
* If i is Prime then add to sum
*/
if(isPrime==1)
{
sum += i;
}
}
printf("Sum of all prime numbers between 1 to %d = %d", n, sum);
return 0;
}