prime number finding from array,c language [duplicate] - c

Question

This program is supposed to print the first x prime numbers, but I noticed that it was printing some non-prime numbers, such as 27 or 35.
I've been looking at it for hours and nothing seems to pop up. So please, if you know what's wrong, tell me.
#include <stdio.h>
int main(){
int i=0, cont=2, prim=2, quant;
printf("Insert number of prime numbers you wish: ");
scanf("%d", &quant);
printf("The first %d prime numbers are:\n", quant);
while(i<quant){
if(prim%cont!=0 && (cont>1 && cont<prim)){
cont++;
}
else if(prim%cont==0 && (cont>1 && cont<prim)){
prim++;
}
else if(prim%cont==0 && cont==prim){
printf("%d\n", prim);
prim++;
cont=2;
i++;
}
}
return 0;
}
UPDATE
Wow, ok, so after 7 years people still stumble upon this question.
In the last 7 years my coding ability has improved somewhat, and now I see how inefficient this program was. Just in case anyone might be trying to find the solution for this and the stumble upon this, here are three solutions much easier and better, and why they work.
Solution 1:
This first solution is very inefficient too, but it works, and is pretty simple to understand. Basically what you have to do is check if each number up to the upper limit is prime or not. To do so, just check if it is divisible by any number up to its square root.
Why its squared root? Because the square root squared is equal to the number, which means that if the number was not divisible up to its square root, it won't be divisible by any number above it, as for it to be divisible by it, it needs to multiply by a smaller number. For example, the squared root of 36 is 6, and 36 is divisible by 9, as 9*4=36. But since 9 is above 6 (which is the squared root), the number that multiplied by 9 gives us 36 is below, and as such, we have already seen that it is not prime, as we have already checked that 36 is divisible by 4. This being said, if no number below the squared root of a number is a natural dividend of the number, than that number is prime.
#include <stdio.h>
int isPrime(int num) {
for (int i = 2; i*i <= num; i++) {
if (num%i==0) {
return 0;
}
}
return 1;
}
void getPrimes(int num) {
int cont = 0;
for (int i = 2; cont < num; i++) {
if (isPrime(i)==1) {
printf("%d\n", i);
cont++;
}
}
}
int main() {
int quant;
printf("Insert number of prime numbers you wish: ");
scanf("%d", &quant);
printf("The first %d prime numbers are:\n", quant);
getPrimes(quant);
return 0;
}
This is inefficient as we are checking if the number is divisible by numbers that won't influence (more on this in the next solution).
Solution 2:
In this second solution, which is more elegant in my opinion, we give use to the Fundamental Theorem of Arithmetic, which states that any number greater than 1 is either a prime number itself, or can be represented by factorizing into the multiplication of prime numbers. This means that if a number is divisible by a non prime number, it is also divisible by the prime numbers that make it up, and thus we only need to check if the number in question is divisible by smaller prime numbers than itself. For this purpose, we store the prime numbers in an array.
#include <stdio.h>
void getPrimes(int num) {
int primes[num];
int cont = 1;
primes[0] = 2;
int current = primes[cont-1]+1;
while (cont < num) {
int before = current;
for (int i = 0; i < cont; i++) {
if (current%primes[i]==0) {
current++;
break;
}
}
if (before == current) {
primes[cont] = current;
cont++;
current++;
}
}
for (int i = 0; i < cont; i++) {
printf("%d ", primes[i]);
}
printf("\n");
}
int main() {
int quant;
printf("Insert number of prime numbers you wish: ");
scanf("%d", &quant);
printf("The first %d prime numbers are:\n", quant);
getPrimes(quant);
return 0;
}
Solution 3:
This final solution is a mix of the two above. Solution 1 checked numbers that weren't prime, having redundancy, and the solution 2 checked numbers above the square root, which can never be a dividend if the smaller numbers aren't. In this solution, we check if the number is divisible only by prime numbers below the squared root.
#include <stdio.h>
void getPrimes(int num) {
int primes[num];
int cont = 1;
primes[0] = 2;
int current = primes[cont-1]+1;
while (cont < num) {
int before = current;
for (int i = 0; (i < cont && primes[i]*primes[i] <= current); i++) {
if (current%primes[i]==0) {
current++;
break;
}
}
if (before == current) {
primes[cont] = current;
cont++;
current++;
}
}
for (int i = 0; i < cont; i++) {
printf("%d ", primes[i]);
}
printf("\n");
}
int main() {
int quant;
printf("Insert number of prime numbers you wish: ");
scanf("%d", &quant);
printf("The first %d prime numbers are:\n", quant);
getPrimes(quant);
return 0;
}
Comparing the execution times of each algorithm for 100 prime numbers, we obtain the following results
Algorithm 1
Algorithm 2
Algorithm 3
0.048 ms
0.068 ms
0.039 ms
Comparing to the algorithm of the original post which takes 0.599 ms, every one of these new solutions is more efficient (the worst being around 10x better), and actually calculates the real values.

Answer 1

Try this
#include<stdio.h>
int prime(int n)
{
int i, j, len=1, brk=0;
int list[200]={2};
for(i=2; i<=n; i++)
{
for(j=0; j<len; j++)
{
if(i%list[j]==0){
brk=1;
break;
}
else
{
brk=0;
}
}
if(brk==0)
{
list[len]=i;
len++;
}
}
for(i=0; i<len; i++)
printf("%d ",list[i]);
}
main()
{
int i, n;
scanf("%d",&n);
prime(n);
}

Answer 2

#PKDOJ If the series is set within 30. You can go blindly with the logic below. But nothing above that series. Adding i%11 clears the 77 as well. Not efficient
if ((i % 2) && (i % 3) && (i % 5)&&(i%7)&&(i%11))

Answer 3

Code:
int count = 0, quant = 5, i, j;
int flag = 0;
for(prim = 2 ; count <= quant ; prim ++) {
flag = 0;
for(j = 2; j < prim/2; j++) {
if(prim % j == 0) {
flag = 1;
break;
}
}
if(flag == 0) {
printf("%d\n", prim);
count++;
}
}
Update your code as:
while(i<quant){
if(cont<prim) {
if(prim%cont!=0) {
cont++;
} else {
prim++;
cont = 2; // restart cont
}
}
else if(prim%cont==0 && cont==prim){
printf("%d\n", prim);
prim++;
cont=2;
i++;
}
}

Answer 4

Simple way is to find is: if a number is not divisible by 2,3 & 5, its a prime number.
#include <stdio.h>
int main()
{
int num = 35;
int i = 5;
printf("1 2 3 5");
while (i <= num)
{
if ((i % 2) && (i % 3) && (i % 5))
{
printf(" %d",i);
}
i++;
}
printf("\n");
return 0;
}