write the prime no program in c++?
Answers were Sorted based on User's Feedback
Answer / md.irfan(rourkela)
#include<iostream.h>
#include<conio.h>
main()
{
int n,i,f=1;
cout<<"enter any no to check prime";
cin>>n;
for(i=2;i<n;i++)
{
if(n%i==0)
k=2;
}
if(k==2)
cout<<"the no is not prime"<<endl;
else
cout<<"the no is prime";
}
getch();
}
Is This Answer Correct ? | 145 Yes | 91 No |
Answer / swapnil
/*To check whethere Entered no. is PRIME or NOT*/
#include<iostream.h>
#include<conio.h>
main()
{
int num,i;
clrscr();
cout<<"Enter the any no.="<<"\n";
cin>>num;
for(i=2;i<=num;i++)
{
if(num%i==0)
{
break;
}
}
if(i==num)
{
cout<<"The number entered is a
PRIME no.";
}
else
{
cout<<"The number entered is NOT a
PRIME no.";
}
getch();
}
Is This Answer Correct ? | 74 Yes | 41 No |
Answer / indu
#include<stdio.h>
#include<conio.h>
main()
{
int n,s=0,i;
clrscr();
printf("Enter the number\n");
scanf("%d",&n);
for(i=2;i<n/2;i++)
{
if(n%i==0)
s++;
}
if(s>=1)
printf("%d is not prime",n);
else
printf("%d is prime",n);
getch();
}
ill execute it .it's cent percent correct.
Is This Answer Correct ? | 40 Yes | 22 No |
Answer / prits
#include <iostream>
using namespace std;
void main()
{
int num;
int flag = 0;
cout<<"enter a number"<<endl;
cin>>num;
for(int i=2;i<num;i++)
{
if((num%i) == 0)
{
flag = 1;
break;
}
}
if (flag == 1)
cout<<"Not Prime"<<endl;
else
cout<<"Prime"<<endl;
}
Is This Answer Correct ? | 32 Yes | 15 No |
Answer / darren chang
bool checkPrime(int input)
{
for(int i=2; i<(input/2); i++)
{
if((input%i)==0)
{
return false;
}
}
return true;
}
Is This Answer Correct ? | 14 Yes | 7 No |
Answer / daljeet singh
#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int n,i;
char prime;
prime='Y';
cout<<"\n enter a number";
cin>>n;
for(i=2;i<=n/2;++i)
{
if(n%i==0)
{
prime='N';
break;
}
}
if(n==1)
prime='N';
if(prime=='Y')
cout<<"\n entered number is prime";
else
cout<<"\n not a prime no";
getch();
}
Is This Answer Correct ? | 10 Yes | 3 No |
Answer / rajan maheshwari
#include<iostream.h>
#include<conio.h>
void main()
{clrscr();
int a,c=0,i=2;
cout<<"Enter A Number = ";
cin>>a;
while(i<=a/2)
{
if(a%i==0)
{c=0;
break;}
else
{i++;
c=1;
}}
if(c==1||a==2||a==3)
cout<<"\nPrime Number";
else
if(a==1)
cout<<"\nNeither Prime Nor Composite";
else
cout<<"\nNot a Prime Number";
getch();
}
Is This Answer Correct ? | 3 Yes | 0 No |
Answer / suresh kumar
#include<iostream.h>
#include<conio.h>
void main()
{
int n,k,i=1;
cout<<"enter any no to check prime";
cin>>n;
for(i=2;i<n;i++)
{
if(n%i==0)
k=2;
}
if(k==2)
cout<<"the no is not prime"<<endl;
else
cout<<"the no is prime";
getch();
}
Is This Answer Correct ? | 4 Yes | 1 No |
Answer / waleed
#include<iostream.h>
#include<conio.h>
int prime(int);
void main()
{
clrscr();
int num,x;
cout<<"Enter the no.=";
cin>>num;
x=prime(num);
if(x==1)
{
cout<<"Number is prime";
}
else
{
cout<<"Number is not prime";
}
getch();
}
int prime(int x)
{
int k;
for(int i=2;i<x;i++)
{
if (x%i==0)
{
k=2;
}
}
if (k==2)
return 0;
else
return 1;
}
Is This Answer Correct ? | 2 Yes | 0 No |
Answer / prateek
I found some good and new ways to write this programe.
Thankx to all. Well even I have also tried to make this
programe. By making use of "while loop";
#include<iostream>
#include<conio.h>
using namespace std;
void main(){
int num,n,i=0,flag=0;;
cout<<"Enter the number";
cin>>num;
n=num/2;
while(i<n)
{
++i;
if(num%i==0 && i!=1)
{
flag=1;
break;
}
else
{
flag;
}
}
if(flag)
{
cout<<"The number is not a prime number";
}
else
{
cout<<"The number is a prime number";
}
getch();
}
Plz notice, that I have divided the number by 2. Suppose
user input 42. The highest divisible value of 42 will be
its half, i.e., 21(21*2=42). So there is no need to check
the loop condition until the value of 'i' reach num. Becoz
it is for sure, that the values more than its half are not
divisible. This will increase the efficiency of the
programe.
Is This Answer Correct ? | 1 Yes | 0 No |
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