Question

How to make program of prime numbers series from 1-100 in C++.

Answer 1

The C Programming Language, Second Edition
by Brian W. Kernighan and Dennis M. Ritchie.
Prentice Hall, Inc., 1988.
ISBN 0-13-110362-8 (paperback), 0-13-110370-9 (hardback).

Answer 2

int n;
cin>>n;
for(int i=2;i<=n-1;i++)
if(n==2)
Cout<<"Prime Number";
else
if(n%i==0)
cout<<"Entered Number is not a prime no";
else
cout<<"prime number";

i think the above code should work out for any given number.try this out!

Answer 3

I am yet to learn C++ but i hope this one written in C should help u-it generates prime numbers from 2 to your specified limit...for your case 100:
#include <stdio.h>
#include <conio.h>
#include <math.h>
int main ()
{
int n,x,y,c,a;
n=2;
while(n<1000)
{
y=1;
a=0;
c=0;

do{
x=sqrt(n);
if((n%y)!=0)
c=0;
else
a=a+1; /*a counts the number of times n is successfully divided*/
y++;
}while(y<=x);

if(c==0&&a==1) /*a=1 makes sure 1 is the only divisor that successfully divides n*/
printf("%i\t",n);
n++;
}

getch();
return 0;
}

Answer 4

This program works by induction, it starts printing 2 and storing 2 as a "known prime" on the array v, then checks if 3 is a multiple of any of the known primes, and if it isn't, 3 is a prime, it's printed, added to the list of known primes, and so on with 4, 5, 6... n was fixed as 100, but you can easily change its value or receive it as input.

#include <iostream>
using namespace std;

int main()
{
unsigned long long n, i, j, cnt;
n=100;
unsigned long long v[n];
cnt=0;

for(i=2;i<=n;i++)
{
for(j=0;j<cnt;j++) //
{if((i%v[j])==0) break;} // if n isn't a multiple of any v[j]
if(j==cnt) //then do this:
{cout<<i<<endl; v[cnt]=i; cnt++;}
}

return 0;
}

Answer 5

all of the above are correct.. but slow.The whole way of thinking is, every number has a divisor smaller or equal than its square root, so you have to loop over the diastema [2,sqrt(n)].This solution has a compexity of (n*sqrt(n)) slightly smaller than the O(n^2) proposed aobve.The significant improvement is to use the following theorem "Every number has at least one prime divisor" so you start with an array containing 2 and you start looping into the diastema [2,p] where p is the limit, once.For every number you find you check if it is divided by any of the numbers into the array, if not then it is a prime number.The pseudocode that implements this idea is the following. stack array[p] for(i=0 to p): for(j=0 to array.size): if(my_number mod array[j]==0) break; end of loop if(j==array.size) array.push(my_number) end of loop export result very brief and not a good code but i think it's kinda clear.If you are using c++ it's a good idea to use the stl stack in order to quickly use elements,otherwise an array would do. I have no idea what the average complexity of the algorithm is, it depends on the number of prime numbers.