val = 2*3*5*7*11
for(i = 0 to n-1)
  if(val%a[i] == 0)
     count++,sum+=a[i];

surender

On Tue, Jul 5, 2011 at 10:10 PM, Rajeev Bharshetty <
rajeev.open.1...@gmail.com> wrote:

> Clarification : The number (count) is the number of elements between 1 and
> n which are not evenly divisible by 5 prime numbers
>                     and the result is the sum of all the numbers between 1
> and n which are not evenly divisible by 5 prime numbers . Right???
>
> For Example : if n=5 then
>                 count ->2 (i.e  1 and 3 )
>               and result = 5
>
> If I am wrong ,Please Correct me and give me the outputs expected
>
> Thank You
>
> Rajeev N B
>
>
> On Tue, Jul 5, 2011 at 12:17 PM, shiv narayan 
> <narayan.shiv...@gmail.com>wrote:
>
>> Write a program that accepts an input integer n, and calculates the
>> number and sum of all the numbers between 1 and n (inclusive) that are
>> NOT evenly divisible by ANY of the first 5 prime numbers (2,3,5,7,11).
>> The program should print out a clearly labeled count and sum
>> my code is : it is not giving correct result
>>
>> #include <iostream>
>> #include<conio.h>
>> using namespace std;
>> int main()
>> {
>> int userInteger = 0;
>> cout << "Enter A Number "<<endl;
>> cin >> userInteger; // Ask For a number from the user
>> if (userInteger > 0) // Is the number valid?
>> {
>> int result = 0;
>> int prime[5] = { 2, 3, 5, 7, 11 };
>> int a = 1, count = 0;
>> while (a < userInteger) // Looping to user's input
>> {
>> int b = 0;
>> while (b < 5) // Looping the prime numbers array
>> {
>> if (a % prime[b])
>> {
>> result += a; // If Not evenly divisible by prime number at index 'b'
>> count++;
>> }
>> b++;
>> }
>> a++; // Increment the counter
>> }
>> cout << "Numbers Not evenly divisible by 5 prime numbers: " << count
>> << endl;
>> cout << "The Sum of numbers not evenly divisible by 5 prime numbers: "
>> << result << endl;
>> }
>> getch();
>> return 0;
>> }
>>
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