# C Program to Print Prime Numbers In A Given Range

## Prime Numbers in a Given Range in C

A number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11).

The C program reduces the number of iterations within the loop. It is made to identify or calculate the prime numbers within a given range of numbers inserted by the user.

Ex:- if the user enters a range as 40 – 50

In that range 41, 43, 47, these three number are prime number. ## Methods Discussed in page

We have discussed the following methods

• Basic checking prime by only checking first n/2 divisors
• Checking prime by only checking first √n divisors
• Checking prime by only checking first √n divisors, but also skipping even iterations.

## Method 1

• Take lower and upper bound values from the user
• Run a loop in the iteration of (i) b/w these bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to next iteration

### Code

Run
```#include <stdio.h>

int checkPrime(int num)
{
// 0, 1 and negative numbers are not prime
if(num < 2){
return 0;
}
else{
// no need to run loop till num-1 as for any number x the numbers in
// the range(num/2 + 1, num) won't be divisible anyways.
// Example 36 wont be divisible by anything b/w 19-35
int x = num/2;
for(int i = 2; i < x; i++)
{
if(num % i == 0)
{
return 0;
}
}
}
// the number would be prime if we reach here
return 1;
}

int main()
{
int a=10, b=20;

for(int i=a; i <= b; i++){
if(checkPrime(i))
printf("%d ",i);
}

return 0;
}
//Time Complexity: O(N^2)
//Space Complexity O(1)```

### Output

`11 13 17 19 `

## Method 2

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has better time complexity of O(√N)

• Take lower and upper bound values from the user
• Run a loop in the iteration of (i) b/w these bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to next iteration

### Code

Run

```#include  <stdio.h>
#include  <math.h>

int checkPrime(int num)
{
// 0, 1 and negative numbers are not prime
if(num < 2){
return 0;
}
else{
// A number n is not a prime, if it can be factored into two factors a & b:
// n = a * b

/*Now a and b can't be both greater than the square root of n, since
then the product a * b would be greater than sqrt(n) * sqrt(n) = n.
So in any factorization of n, at least one of the factors must be
smaller than the square root of n, and if we can't find any factors
less than or equal to the square root, n must be a prime.
*/
for(int i = 2; i < sqrt(num); i++)
{
if(num % i == 0)
{
return 0;
}
}
}
// the number would be prime if we reach here
return 1;
}

int main()
{
int a, b;
a=10,b=20;

for(int i=a; i <= b; i++){
if(checkPrime(i))
printf("%d ",i);
}

return 0;
}
//Time Complexity: O(N√N)
//Space Complexity O(1)

// This method is obviously  faster as has better time complexity
```

### Output

```Enter the lower and upper ranges:
20 40
23 25 29 31 37```

## Method 3

The outer logic remains the same. Only the method to check prime changes to make code more optimized. Has same time complexity of O(√N).

But makes around half lesser checks

• Take lower and upper bound values from the user
• Run a loop in the iteration of (i) b/w these bounds.
• For each, i check if its prime or not using function checkPrime(i)
• If i is prime print it else move to next iteration

### Code

Run
```#include <stdio.h>
#include <math.h>

int checkPrime(int n)
{
// 0 and 1 are not prime numbers
// negative numbers are not prime
if (n <= 1)
return 0;

// special case as 2 is the only even number that is prime
else if (n == 2)
return 1;

// Check if n is a multiple of 2 thus all these won't be prime
else if (n % 2 == 0)
return 0;

// If not, then just check the odds
for (int i = 3; i <= sqrt(n); i += 2)
{
if (n % i == 0)
return 0;
}

return 1;
}

int main()
{
int a, b;
a=10,b=20;

for(int i=a; i <= b; i++){
if(checkPrime(i))
printf("%d ",i);
}

return 0;
}
//Time Complexity: O(N√N)
//Space Complexity O(1)

// This method is obviously faster as makes around half lesser comparision due skipping even iterations in the loop

```

### Output

```Enter the lower and upper ranges:
30 50
31 37 41 43 47```

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### One comment on “C Program to Print Prime Numbers In A Given Range”

• #include

int main()
{
int i,num,count=0,j;
scanf(“%d”,&num);
for(i=1;i<=num;i++){
count=0;
for(j=1;j<=i;j++){
if(i%j==0){
count++;
}
}
if(count==2){
printf("it is a prime no:%d\n",i);

}
}

return 0;
} 0