help me please?! prime number in c++?

Question

is it possible to write the same program wit same commands without flag and also please describe flag

#include


int main(void){
int j,n,i,flag;
printf("Please enter a number : ");
scanf("%d",&n);
if(n<2) {
printf("\nNo prime numbers available");
return 1;}
printf("Thank you : 2");
for(i=2;i<=n;i++){
for(j=2;j<=i-1;j++){
if(i%j==0){
flag=0;
break;}
flag=1;
}
if(flag==1)
printf("%5d",i);
}

return 0;
}

Answer 1

u need atleast one flag variable. I have reduced ur code. wat i have given is very simple code.

please correct the syntax. but the logic is correct.

flag=1;

for(i=2;i if(n%i==0){
flag=0;
break;
}
}
if(flag==1){
printf("prime");
}
else
printf("not a prime");


//Note:-
/*
n is the number for which it has to be checked for prime.
flag=1 (initially assuming that the given number is prime and reinitializing to 0 after checking the condition in if loop.
*/

Answer 2

Yes. Instead of for loops you should be using while loops. The flafg here is only used for breaking the loop. You will not need it if you use while loop.

i =2;
while(i<=n)
{
j=2;
while (j<=i-1)
{
if (I%j++ == 0)
{
break; // this will break the loop for j
}
}// end of while for j

if (j == i) printf("%d ",i);
i++;
} //end of while for i
}

Answer 3

//try this
#include
void main()
{
int n,i,flag;
flag=0;
printf("Enter the Number:\t\t");
scanf("%d",&n);
if(n<2)
printf("\n%d is not prime",n);
else
{
for(i=0;i {
if(n%i)
flag=1;
}

if(flag)
printf("%d is not prime",n);
else
printf("%d is prime",n);
}
getch();
}