I'm trying to make a program that checks whether a given number is a prime number or not. However, my program just gives me the 2 time table, and I don't know why.
Here is my main class:
#include <stdio.h>
#include "defs.h"
#include "checkprime.c"
int Prime[MaxPrimes];
int main()
{
int UpperBound;
int N;
int *ba = &UpperBound;
printf("enter upper bound\n");
scanf("%d",ba);
Prime[2] = 1;
for (N = 3; N <= *ba; N+= 2)
{
CheckPrime(N);
if (Prime[N]== 1) printf("%d is a prime\n",N);
}
}
Here is my checkprime.c
#include "defs.h"
#include "externs.h"
int CheckPrime(int K)
{
int J;
J = 2;
while (1)
{
if (Prime[J] == 1)
{
if (K % J == 0)
{
Prime[K] = 0;
return 0;
}
J++;
}
break;
}
Prime[K] = 1;
}
There are some problems in CheckPrime with the loop exit conditions. Use the following instead:
int CheckPrime(int K)
{
int J;
for (J=2; J*J <= K; J++) {
if (Prime[J] == 1) {
if (K % J == 0) {
Prime[K] = 0;
return 0;
}
}
}
Prime[K] = 1;
return 1;
}
The rest of it should work with this change.
Related
I have make this program that calculates number factorization such as 60 = 2^2 * 5 * 3.
How can i modify my code in order to print POWERFUL NUMBERS such as 9000 = 2^3 * 3^2 * 5^3 without using math.h library and without using arrays?
Thank you very much!!
#include<stdio.h>
#define MAX 1000
int main(){
int num;
int counter;
int number;
char factorizationOutput;
int isAchiles = 0;
int factor=2;
for(counter=2;counter<=MAX;counter++){
isAchiles = 1;
number=counter;
int factor=2;
while(factor<number){
int power=0;
if(number%factor==0){
while(number%factor==0){
number=number/factor;
power++;
}
if(power == 1){
isAchiles = 0;
}
printf("%d^%d",factor,power);
if(number!=1)
printf(" X ");
}
factor++;
}
if(number!=1)
printf("%d^1.\n",factor);
if(isAchiles == 1){
printf("factorazation of number %d is: ",counter);
}
}
}
#include<stdio.h>
int main(void)
{
int n;
scanf("%d", &n);
printf("%d = ", n);
for(int i = 1; i <= n; i++)
{
int count = 0;
for(int j = 1; j <= i; j++)
{
if(i % j == 0)
{
count++;
}
}
int l = 0;
if(count == 2)
{
while(n % i == 0)
{
l++;
n = n/i;
}
printf("%d^%d*", i, l);
}
}
}
I've been trying to do the Love-Letter Mystery Challenge on Hackerrank.
Here are the rules: https://www.hackerrank.com/challenges/the-love-letter-mystery
And here's my solution:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAX 1000
int check_palindrome(char *A)
{
int i = 0, j;
while(A[i])
i++;
i -= 1;
for(j = 0; j <= i; j++)
{
if(A[j] != A[i - j])
return 0;
}
return 1;
}
int love_letter(char *A)
{
int i = 0;
int j;
int times;
while(A[i])
i++;
i -= 1;
if(i == 0)
return 0;
if(check_palindrome(A))
return 0;
for(j = i; j >= 0; j--)
{
while(A[j] != 'a')
{
if(check_palindrome(A))
return times;
else
{
A[j] -= 1;
times += 1;
}
}
}
return times;
}
int main() {
int t, i;
char a[MAX];
scanf("%d", &t);
for(i = 0; i < t; i++)
{
scanf("%s", a);
printf("%d\n", love_letter(a));
}
return 0;
}
While testing in on my computer, I get the right output. But, when I try to run the code on Hackerrank, it tell's that my program always gives an output of:
0
0
0
0
That's wrong of course, and it fails the testcase. But why is that? Is there something different about C or something? Or is it just a problem with the site? Or with my code?
At a minimum, you seem to have forgotten to initialize the variable "times".
In terms of the actual algorithm, keep in mind that to make the letters match, you can decrement either (or both) of them. I don't think you handle all cases properly.
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;
}
This is my code which is supposed to output prime numbers only.
#include <stdio.h>
int prime(int n){
int j;
for (j=2;j<=n/2;j++){
if((n%j)==0){
return 0;
}
else{
return 1;
}
}
}
void main(){
int i,p;
for (i=2;i<=100;i++){
p=prime(i);
if(p==1){
printf("%d \n",i);
}
}
}
The result is 2,3,7,9,11,13,15....
not 2,3,5,7,11,13....
What did I do wrong?
Your probably want:
int prime(int n){
int j;
for (j=2;j<=n/2;j++)
if((n%j)==0)
return 0;
return 1;
}
Prime Numbers are numbers which are only divisible by two numbers. eg 2,3,5,7,11 etc.
int main()
{
int i,j,n=0;
for(i=2;i<=100;i++)
{
for(j=1;j<=i;j++)
{
if(i%j==0)
{
n++;
}
}
if(n==2)
printf("%d\n",i);
n=0;
}
getch();
}
Try this :
int prime(int n){
int j;
int isPrime = 1;
for (j=2;j<=n/2;j++){
if((n%j)==0){
isPrime = 0;
break;
}
}
return isPrime;
}
If you're trying to find out Prime numbers up to a given number(in your case 2..100) building a prime table will speed up the process:
NOTE: it's important that inputs are sequential to help build prime table.
#include <stdio.h>
#include <math.h>
/* Building a prime table for fast look up
* NOTE : Numbers should be input from 2 to n in a sequential pattern to help build prime table; */
# define PRIME_TAB_MAX 200
int prime_specific(int num)
{
static int primeTab[PRIME_TAB_MAX] = { 0, };
static int primeCount = 0;
int check_limit ;
unsigned int idx;
if (num < 2)
return 0;
if (primeCount > 0) {
check_limit = (int)sqrt(num);
for (idx = 0; idx < primeCount && primeTab[idx] <= check_limit; idx++) {
if (0 == (num % primeTab[idx])) {
return 0;
}
}
}
else {
for (idx = 2; idx <= num / 2; idx++)
if (0 == (num % idx))
return 0;
}
if (primeCount < PRIME_TAB_MAX) {
primeTab[primeCount++] = num;
return 1;
}
else {
fprintf(stderr, "\nERROR: Prime Table is full");
return (-1); //Internal error
}
}
int main(void)
{
unsigned int idx;
int status ;
for (idx = 2; idx <= 1000; idx++) {
status = prime_specific(idx);
switch (status)
{
case 1:
fprintf(stderr, "%d\n", idx);
break;
case 0:
//Do nothing
break;
default:
fprintf(stderr, "\nERROR:Internal Error");
return (-1);
}
}
return 0;
}
Code:
#include <stdio.h>
int main ()
{
/* local variable definition */
int i, j;
for(i=2; i<100; i++)
{
for(j=2; j <= (i/j); j++)
if(!(i%j)) break; // if factor found, not prime
if(j > (i/j)) printf("%d is prime\n", i);
}
return 0;
}