[algogeeks] help to code

2011-07-05 Thread shiv narayan 
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
#includeconio.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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Re: [algogeeks] help to code

2011-07-05 Thread Rajeev Bharshetty 
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.comwrote:

 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
 #includeconio.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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Re: [algogeeks] help to code

2011-07-05 Thread surender sanke 
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.comwrote:

 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
 #includeconio.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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Re: [algogeeks] help to code

2011-07-05 Thread Tushar Bindal 
I think you are getting it wrong.
Surender, your solution says that numbers divisible by all of the first 5
prime numbers will be taken into account whereas the question says that only
the numbers *not* evenly divisible by *any* of the first 5 prime numbers are
to be added.

Shiv,
you are making two mistakes:
1. you say n is also included but your code excludes it.
2. you are just checking if a number is not divisible by one of the 5 prime
numbers. If it is not divisible by any one number, you add it to your
result, whereas your language of the question suggests that it should be
that numbers which are not divisible by any of the five numbers, i.e., not
divisible by all the 5 numbers ( not the ones divisible by just ant one of
the 5).

Hope I am correct in my interpretation.

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Re: [algogeeks] help to code

2011-07-05 Thread Tushar Bindal 
If my interpretation is right, following should be the code.


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,b, count = 0;
*for(a=1;a=userInteger;++a) // Looping to user's input
{
  for(b=0;b  5;++b) // Looping the prime numbers array
   {
   if (a % prime[b] == 0) //if divisible by any number, just end it there
   break;
  }
 //break or end of loop will bring the control here
 if(b==5) //a was not divisible by any of the first 5 prime numbers
 {
 result+=a;
 ++count;
 }
}*
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;
}



As per my solution,
Test Cases:
1)
userInteger = 20
count = 4
Numbers will be: 1, 13, 17, 19
result = 50

2)
userInteger = 35
count = 7
Numbers will be: 1, 13, 17, 19, 23, 29, 31
result = 133




Even numbers can never be there in this list as they are all divisible by 2.
Bfore 169, only prime numbers can be included.
Hope my interpretation of your question was correct

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