I want to calculate the average of the prime numbers between 1 to 10 and I have written a program which is as follows:
#include <stdio.h>
int main()
{
int i, j, sum = 0, count = 0;
loop1:
for(i = 2; i <= 10; i++)
{
for(j = i - 1; j > 1; j--)
{
if(i % j == 0)
{
goto loop1;
}
}
sum = sum + i;
count++;
}
printf("The avg:%d", (sum / count));
return 0;
}
Please help me whether the program is correct.
Simple answer for you:
#include <stdio.h>
int main() {
float sum = 0, count = 0, average;
for(int i=2; i<11; i++){
for(int j=2; j<=i; j++){
if(j==i){
sum+=i;
count++;
}else if(i%j==0){
break;
}
}
}
average=sum/count;
printf("Average = %.2f", average);
return 0;
}
Please ensure you do not make a habit of usinggoto: statements. Your program is incorrect. Here, when the statement
if(i % j == 0)
returns true, thegoto: statement takes control to the beginning of the parent for loop and the loop will run from start again. This way, you will never get your desired output. As per your question, your solution is wrong. A modified approach is
#include <stdio.h>
int main()
{
int i, j, sum = 0, count = 0,flag;
for(i = 2; i <= 10; i++)
{
flag=0;
for(j = i - 1; j > 1; j--)
{
if(i % j == 0)
{
flag=1;
break;
}
}
if(flag==0)
{
sum = sum + i;
count++;
}
}
printf("The avg:%d", (sum / count));
return 0;
}
What do you mean by "Correct" ?
If correct to you means that it works: well that's easy to verify for yourself.
If correct means that you are hitting the best practices, well then, nope, you missed them I'm afraid.
goto: about the most easy of all "do not use that" signs.
A "better" approach would be to write a function that tests if a number is a prime or not.
Related
I am a beginner in C and was working on a program that can output the prime numbers within the range using Eratosthenes' sieve. However, the program does not output anything and is in a loop, I believe this has to do with the repetitive use of for statements in my code but also have no idea where I may be going wrong. I tried putting break statements with if statements but it seems like it does not work. Thank you in advance.
#include <stdio.h>
#include <math.h>
int sieve(int limit);
int main() {
int maximum = 0;
printf("Enter the limit of prime numbers:");
scanf_s("%d", &maximum);
sieve(maximum);
return 0;
}
int sieve(int limit) {
int array[100];
int common = 2;;
for (int i = 0; i <= limit; i++) {
array[i] = i + 1;
}
for (;;) {
for (int j = 0; j <= limit; j++) {
if (array[j] % common == 0 && !(array[j] == common)) {
array[j] = 0;
array[0] = 0;
}
}
for (int k = 0; k <= limit; k++) {
if (!(array[k] == 0) && !(common == array[k])) {
common = array[k];
break;
}
}
if (common >= sqrt(limit))
break;
}
for (int o = 0; o < limit; o++) {
if (!(array[o] == 0)) {
printf("%d ", array[o]);
}
}
return 0;
}
I hope I'm not out of my depth here since it's been a while since I have touched C.
You could set a boolean variable to false. Within the child loop you set it to true. Add an if statement within the parent loop to check if that variable is true and break the parent loop if it is.
Quick pseudo-code:
bool t = false
parent loop:
child loop:
if something
t = true
if t
break
The outer loop runs for ever because you do not increment common inside the loop body.
Note also that you can enumerate all multiples of common with a simple addition instead of a costly % operator.
Here is a modified version:
#include <stdio.h>
int sieve(int limit);
int main() {
int maximum = 0;
printf("Enter the limit of prime numbers:");
scanf_s("%d", &maximum);
sieve(maximum);
return 0;
}
int sieve(int limit) {
int array[limit + 1];
for (int i = 0; i <= limit; i++) {
array[i] = i;
}
for (int common = 2; common * common <= limit; common++) {
// remove all multiples of common. Start at common*common
// because lesser multiples have already been removed as they
// are multiples of a smaller number
for (int i = common * common; i <= limit; i += common) {
array[i] = 0;
}
}
for (int i = 1; i <= limit; i++) {
if (array[i] != 0) {
printf("%d ", array[i]);
}
}
printf("\n");
return 0;
}
Note that you can use an array of unsigned char initialized to 1 and print i instead of array[i].
I was wondering, how can I add up the divisors displayed once I run my code?
#include <stdio.h>
#include <stdlib.h>
int main()
{
int n, i;
scanf("%d", &n);
for(i = 1; i < n; i++)
{
if(n % i == 0)
{
printf("%d\n", i);
}
}
return 0;
}
If I were to enter, say, 25, it would print out 1, 5. I was wondering how to add up the two numbers 1 and 5?
Could it be as simple as this? You will want to use a simple increment operator (+=), to increment the variable sum.
int main(void)
{
int n, i, sum = 0;
if( scanf("%d", &n)!= 1){
fprintf(stderr,"Error in input\n");
exit(EXIT_FAILURE);
}
for(i = 1; i < n; i++)
{
if(n % i == 0)
{
printf("%d\n", i);
sum += i;
}
}
printf("Sum of divisors: %d\n", sum);
return 0;
}
How to add up divisors of an integer n?
Iterating n times as with for(i = 1; i < n; i++) can take a long time when n is some high value, especially if code used 64-bit integers. Instead only iterate sqrt(n) times - much faster
int factor_count(int number) {
if (number <= 1) {
return TBD_Code(number); // OP needs to define functionality for these cases.
}
int sum = 1;
int quotient;
int divisor = 1;
do {
divisor++;
quotient = number/divisor;
int remainder = number%divisor;
if (remainder == 0) {
sum += divisor;
if (quotient > divisor) sum += quotient;
}
} while (divisor < quotient);
return sum;
}
Additional improvements noted here.
I have written some code to ask the user for n, then print the prime numbers up to n. However when I use it, i.e with 10, it only prints the non-prime numbers
/* Asks for the amount of prime numbers you would like to print, then prints them */
#include <stdio.h>
int main(void)
{
int n, i, j, check;
printf("How many prime numbers would you like to print? ");
scanf("%d", &n);
for (i = 2; i <= n; i++) {
check = 0;
for (j = 2; j < i ; j++) {
if (i % j == 0) {
check = 1;
if (check == 1) {
printf("%d\n", i);
}
}
}
}
return 0;
}
How many prime numbers would you like to print? 10
4
6
6
8
8
9
10
10
I've tried everything but I think I am missing something really trivial!
This is how it should be:
for (i = 2; i <= n; i++)
{
check = 0;
for (j = 2; j < i ; j++)
{
if (i % j == 0)
{
check = 1;
break;
}
}
if (check == 0)
{
printf("%d\n", i);
}
}
Also, in the inner loop you don't have to divide the number till j < i. You don't have to go beyond i/2.
As Weather Vane said, the mod operator % returns 0 if i is exactly divisible by j and if this is true then the number is not prime. Your conditional statement is backwards.
#include <stdio.h>
int main(void)
{
int n, i, j, check;
printf("How many prime numbers would you like to print? ");
scanf("%d", &n);
for (i = 2; i <= n; i++)
{
check = 0;
for (j = 2; j < i ; j++)
{
if (i % j == 0)
{
check = 1;
break;
}
}
if (check == 0)
{
printf("%d\n", i);
}
}
return 0;
}
How many prime numbers would you like to print? 10
2
3
5
7
Several problems.
First, when you set check = 1, that means that i divides evenly, so n is not prime, so you shouldn't print it. You should be printing the number when check == 0.
Second, you're printing each time through the inner loop. You should test check at the end of the loop, to ensure that none of the numbers divided it.
As an improvement, there's no need to keep checking once you find one number that divides evenly. So you can break out of the inner loop as soon as you set check = 1.
#include <stdio.h>
int main(void)
{
int n, i, j, check;
printf("How many prime numbers would you like to print? ");
scanf("%d", &n);
for (i = 2; i <= n; i++) {
check = 0;
for (j = 2; j < i ; j++) {
if (i % j == 0) {
check = 1;
break;
}
}
if (check == 0) {
printf("%d\n", i);
}
}
return 0;
}
try looking at this code
#include <stdio.h>
int IsPrime(int num)
{
int i = 2;
for (i = 2; i < num; i++) if (num % i == 0) return 0;
return 1;
}
int main(void)
{
int n, i;
char *nStr = (char*)malloc(10);
printf("How many prime numbers would you like to print? ");
fgets(nStr, 9, stdin);
n = atoi(nStr);
for (i = 1; i <= n; i++) if (IsPrime(i)) printf("%d\n", i);
getchar();
return 0;
}
and about your code, you should print the number only if check remains 0.
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}
I want to print this pattern like right angled triangle
0
909
89098
7890987
678909876
56789098765
4567890987654
345678909876543
23456789098765432
1234567890987654321
I wrote the following code:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
for(j=1;j<=f;j++,k--)
{
k=i;
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
for(x=1;x<f;x++,z--)
{
z=9;
printf("%d",z);
}
printf("%d/n");
}
getch();
}
What is wrong with this code? When I check manually it seems correct but when compiled gives different pattern
Fairly simple: use two loops, one for counting up and one for counting down. Print literal "0" between the two.
#include <stdio.h>
int main()
{
for (int i = 0; i < 10; i++) {
for (int j = 10 - i; j < 10; j++)
printf("%d", j);
printf("0");
for (int j = 9; j >= 10 - i; j--)
printf("%d", j);
printf("\n");
}
return 0;
}
Like H2CO3's, but since we're only printing single digits why not use putchar():
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int i, j;
for(i = 0; i < 10; ++i)
{
// Left half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
// Center zero.
putchar('0');
// Right half.
for(j = 0; j < i; ++j)
putchar('9' - i + j + 1);
putchar('\n');
}
return EXIT_SUCCESS;
}
Modified Code:
Check your errors:
# include<stdio.h>
# include<conio.h>
int main()
{
// clrscr();
int i,j,x,z,k,f=1;
for ( i=10;i>=1;i--,f++)
{
k=i; // K=i should be outside of loop.
for(j=1;j<=f;j++,k++)
{
if(k!=10)
{
printf("%d",k);
}
if(k==10)
{
printf("0");
}
}
z=9; //z=9 should be outside loop.
for(x=1;x<f;x++,z--)
{
printf("%d",z);
}
printf("\n");
}
//getch();
return 0;
}
You are defining k=i inside the for loop(loop which has j) so every time k gets value of i and thus it always get value of i and prints that value and your another condition(if(k==10)) will never be true because every time k takes value of i and i is less than 10 after first iteration of loop and z=9 inside loop so every time loop is executed it is taking value z=9 so it is printing wrong value.
Here's a C# version:
static void DrawNumberTriangle()
{
for (int line = 10; line >=1; line--)
{
for (int number = line; number < 10; number++)
{
System.Console.Write(number);
}
System.Console.Write("0");
for (int number = 9; number > line - 1; number--)
{
System.Console.Write(number);
}
System.Console.WriteLine();
}
}
I'd suggest renaming your i,j,x,z,k,f variables to ones that have meaning like the one's I used. This helps making your code easier to follow.
Rather than output the mid 0 using printf, why not print it using the loops itself.
The following short and simple code can be used:
int main()
{
int m = 10, n, p;
while(m >= 1)
{
for(n = m; n <= 10; n++)
printf("%d", n % 10);
for(p = n - 2; p >= m; p--)
printf("%d", p );
printf("\n");
m--;
}
return 1;
}
For high throughput (though of questionable merit in terms of clarity):
#include <stdio.h>
int main() {
char const digits[] = "1234567890";
char const rdigits[] = "9876543210";
for (int i = 0; i < 30; ++i) {
int k = i % 10;
fputs(digits + 9 - k, stdout);
for (int j = 9; j < i; j += 10) fputs(digits, stdout);
for (int j = 9; j < i; j += 10) fputs(rdigits, stdout);
fwrite(rdigits, 1, k, stdout);
fputs("\n", stdout);
}
}
#include <stdio.h>
void print(int i){
if(i == 10){
putchar('0');
return ;
} else {
printf("%d", i);
print(i+1);
printf("%d", i);
}
}
int main(void){
int i;
for(i = 10; i>0; --i){
print(i);
putchar('\n');
}
return 0;
}