Examples of C part 23 all prime number till given number
#include<stdio.h> main() { int num,i,count=0,j; printf("Enter a number \n"); scanf("%d",&num); printf("The Prime numbers upto %d are\n",num); for(i=1;i<=num;i++) { count=0; for(j=1;j<=i;j++) { if(i%j==0) { count++; } } if(count==2) { printf("%d\n",i); } } }
Enter a number
30
The Prime numbers upto 30 are
2
3
5
7
11
13
17
19
23
29Explanation:
- The Program starts with initializing
- num → For storing the number to get all prime numbers between 1 and num
- i,j → temporary variables
- count → To count the number of divisors
printf("Enter a number \n"); scanf("%d",&num);
- The loop iterates from i=1 to 8 one by one.Then count is made zero for every iteration.
- Now j will iterate from 1 to i and checks if any prime number is there between 1 and i. For example: if it is 5th iteration then i =5
- Iteration 1: Then the for loop j starts as j=1,j → true then loop continues
- if i%j → 5%1==0 which is true,
- then count=1
- Iteration 2: Then the for loop j starts as j=2,j → true then loop continues
- if i%j → 5%2==0 which is false,
- then count remains same i.e. count=1;
- Iteration 3: Then the for loop j starts as j=3,j → true then loop continues
- if i%j → 5%3==0 which is false,
- then count remains same i.e. count=1;
- Iteration 4: Then the for loop j starts as j=4,j → true then loop continues
- if i%j → 5%4==0 which is false,
- then count remains same i.e. count=1;
- Iteration 5: Then the for loop j starts as j=5,j → true then loop continues
- if i%j → 5%5==0 which is true,
- then count=2
- Iteration 6: Then the for loop j starts as j=6,j → false then loop terminates
- Now if count==2 → true then this will be the prime number so this number is printed i.e i=5 is printed.
- Iteration 1: Then the for loop j starts as j=1,j → true then loop continues
- Like the above iteration it checks each and every value of i whether it is prime or not and if it is prime it will print else not.