1.

Solve : C++ problem?

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i made a program to GET the factors for a NUMBER but i get weird answers back such as Quote

20 has these factors 3999008,3999008,0,0,0,0
for an input of 20.
the code is as followsCode: [Select]#include <cstdlib>
#include <iostream>

using namespace std;
void show( int *arr ,int num );

int main(int ARGC, char *argv[])
{
int num=0;
int numfactor=0;
int numarry=0;
char error[20];
cout << "what number do you want to factor: ";
while (!(cin >> num)) {
cout <<"try again";
cin >> error;
}
for (int i=1;i<=num;i++)
if (!(num % i))
numfactor++;
int * factor =new int [numfactor];
for (int i=1;i<=num;i++)
if (!(num % i)) {
cout <<i<<"\n";
factor[numarry];
numarry++;
}
cout <<num << " has these factors ";
show(factor,numfactor);
cout <<"\n";
system("PAUSE");
return EXIT_SUCCESS;
}
void show (int arr[] ,int x)
{
cout <<arr[0];
for (int i=1;i < x;i++)
cout <<","<<arr[i];
}try using longin what variablemy bad, thought you were doing factorials...(but still better to use long in case you want to find factors of very big numbers)

you are using c++ and std namespace, so use endl instead of "\n". also you did not assign values to your factors array
Code: [Select]using namespace std;
void show( int *arr ,int num );

int main(int argc, char *argv[])
{
int num=0;
int numfactor=0;
int numarry=0;
char error[20];
cout << "what number do you want to factor: ";
while (!(cin >> num)) {
cout <<"try again";
cin >> error;
}
for (int i=1;i<=num;i++){
if (!(num % i))
numfactor++;
}
int * factor =new int [numfactor];
for (int i=1;i<=num;i++)
if (!(num % i)) {
cout <<i<<endl;
factor[numarry]=i;
numarry++;
}
cout <<num << " has these factors ";
show(factor,numfactor);
cout <<endl";
return 0;
}
void show (int arr[] ,int x)
{
cout <<arr[0];
for (int i=1;i < x;i++)
cout <<","<<arr[i];
}

thank you THOUGH i have have two questions
1) why should i use endl instead of /n
2)is this the most efficient method of getting factorsyou can see some discussion here and hereQuote from: mat123 on January 06, 2011, 12:08:30 AM
1) why should i use endl instead of /n
"\n" is literal, and while compilers will generally condense all the instances of a single literal into one, you shouldn't assume anything as far as the compiler goes. endl is a constant, already defined specifically for you to use with iostream.

Quote
2)is this the most efficient method of getting factors

No.

your iterating from 1 to the number, by one every time. first- one will be a factor of any whole number, 2, there is no need to go past the original number divided by 2, since no number B that is larger then Number A/2 will be a factor of A. Also, consider the numbers 2,3,5,7 and 9. if 2 is not a factor, then you KNOW that no multiple of 2 (4,6,8, etc) is a factor either. same with 3 and it's multiples. And that is just for starters.



OK thank you for your help


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