C Program to Find Prime Number between 1 to 100

Prime Number between 1 to 100 in C

Here, on this page, we will discuss the program to find the prime numbers between 1 to 100 in C.

Generally this program is asked as Write a Program to print Prime Numbers from 1 to 100 in C

Prime number

Methods Discussed in page

We have discussed the following methods

  • Basic checking prime by only checking first n
  • 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

  • Set lower bound = 1, upper bound = 100
  • 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

#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 = 1, b = 100;
    
    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

2 3 4 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

Method 2

  • Run a loop in the iteration of (i) b/w 1 and 100 bounds.
  • For each, i check if its prime or not using function checkPrime(i)
  • If i is prime print it else move to the next iteration

Code

#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()
{
    
    for(int i=1; i <= 100; i++){
        if(checkPrime(i))
            printf("%d ",i);
    }
 
    return 0;
}
//Time Complexity: O(N^2)
//Space Complexity O(1)

Output

2 3 4 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

Method 3

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)

  • Run a loop in the iteration of (i) b/w 1 and 100 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

#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()
{
    
    for(int i=1; i <= 100; 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

2 3 4 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

Method 4

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

  • Run a loop in the iteration of (i) b/w 1 to 100 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

#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()
{
    
    for(int i=1; i <= 100; 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

2 3 4 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

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One comment on “C Program to Find Prime Number between 1 to 100”


  • AVN

    easy one
    #include
    int main()
    {
    int i,j,cnt=0,c;
    for(i=2;i<=100;i++)
    {
    for(j=1;j<=i;j++)
    {
    if(i%j==0)
    c=cnt++;

    }
    if(c<2)
    printf("%d\n",i);
    cnt=0;
    }
    }