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How to find smallest n digit prime number as well as largest?
5 ответов
+ 2
Please show your attempt.
+ 3
#include <stdio.h>
int main()
{
int i, Num, prime,largestprime;
for(Num = 100; Num <1000 ; Num++)
{
prime = 0;
for (i = 2; i <= Num/2; i++)
{
if(Num%i == 0)
{
prime++;
break;
}
}
if(prime == 0)
{
if(Num>100){
printf("the smallest three digit prime number is:%d ",Num);
break;
}
}
}
return 0;
}
Here i found smallest using break... For largest what can i do
+ 1
#include <stdio.h>
#define smallest_tested 100
#define largest_tested 1000
int test_for_prime(int some_number){
// Going to loop through all numbers, can be slow if we are testing large numbers
for (int i = 2; i <= some_number/2; i++){ //placed int i in the loop no reason to create a global variable.
if(some_number % i == 0){
return 1;
}
}
return 0;
}
int main(void){
// removed two variable I dont think are needed as globals.
int smallest_prime,largest_prime;
smallest_prime = largest_tested; //initilizing them at opposites to allow greater than less than test to work.
largest_prime = smallest_tested;
// looks like you are only testing 100 to 1000 so I will stick with that
for(int number_to_test = smallest_tested; number_to_test < largest_tested ; number_to_test++){
if(test_for_prime(number_to_test) == 0){
if(number_to_test < smallest_prime){
smallest_prime = number_to_test;
}
if (number_to_test > largest_prime){
largest_prime = number_to_test;
}
}
}
printf("The smallest prime number between 100 and 1000 is %i, the largest prime number between 100 and 1000 is %i\n", smallest_prime, largest_prime);
return 0;
}
Now its your turn change it so you get a min and maximum number to test not just 100 to 1000.