Question

Write a C++ prgm to read a number and print its Factors?

Answer 1

@shadowfiend @walne @myininaya @TuringTest

Answer 2

i want this in simple not complicated...

Answer 3

Sorry. I'm not the person to ask about programming. :(

Answer 4

ok...can to invite some one u knw..

Answer 5

Well here can't you just write a loop from i->num checking the modulii of i and num - i. If the remainder is 0, those are factors. Break out of the loop after i > num / 2. Just off the top of my head here...

Answer 6

IE:
for(i->n)
{
if((i * (num - i)) % num == 0)
i and num - i are factors
if(i > num / 2)
break
}

Something like that.

Answer 7

wat..is this.....i need this in simple.....not complicated..

Answer 8

I'm not sure how much simpler it can get, sorry :/

Answer 9

no..prob..thxxx for helping!

Answer 10

This actually won't work, you'll need to nest two loops and run the multiplication against them to see if the remainder is 0. You can bail once the product exceeds the number.

Answer 11

here is the program.i dont know c++ so well,thats why i posted it in c.....
main()
{
int n,value;
printf("enter the number");
scanf("%d",&n);
if(n<0)
printf("the factorial is none");
else
if(n==0)
printf("the factorial is 1");
else
{
value=factorial(n);//function use
printf("factorial of %d=%d",n,value);
}
}

factorial (int k)
{
int fact=1;
if(k>1)
fact=k*factorial(k-1);//recursion use
return fact;
}

Answer 12

it....c prgm....i want c++

Answer 13

pal i dont know c++ so well,i give u the concept in c,can u kindly write it in c++ and then run it...i hope it will run.......

Answer 14

@hartnn @.Sam. @mukushla

Answer 15

lol...i dont know c++..

Answer 16

ok.....@hartnn u knw...

Answer 17

@Compassionate @sasogeek

Answer 18

i can't write that in c++ cos i don't code c++, but that's a good idea to keep me writing some code, i'll try that with python :)

Answer 19

I don't know much about C++. I only took HTMLx

Answer 20

@A.Avinash_Goutham

Answer 21

#python code

def factorsOf(number):
factors = []
for i in xrange(1,number):
if number%i == 0:
factors.append(i)
return factors

Answer 22

i dont knw thus....))

Answer 23

i need this in a simple...language..))

Answer 24

if you know c++, you can simply transfer the idea... python is quite simpler than c++ though

Answer 25

@theyatin

Answer 26

@Hashir

Answer 27

@ajprincess @shubhamsrg @experimentX @nick66 @eseidl

Answer 28

leme try
int a;
cin>>a;
for (int =1;i<=a ;i++)
{if (a % i == 0)
cout<<i;
}'
hope this helps..

Answer 29

for (int =1;i<=a ;i++)????

Answer 30

the for loop..you must have studies this..

Answer 31

No..lol

Answer 32

even while can be used..

Answer 33

can u write this with out loop

Answer 34

well you gotta use some loop here,,otheewise i dont think will be simple..

Answer 35

just..try...lol

Answer 36

i give up ! ;)

Answer 37

oh...no..

Answer 38

PLZZ....Its urgent

Answer 39

loops is very easy,,comeon!!

Answer 40

thxxx

Answer 41

70 club...

Answer 42

it tells error @shubhamsrg

Answer 43

70 club?
and what code you wrote ?

Answer 44

nothing

Answer 45

it tells undefined symbol i

Answer 46

eh ?
please copy paste your entire code

Answer 47

same error

Answer 48

the code please..??

Answer 49

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int a;
cin>>a;
for (int =1;i<=a ;i++)
{if (a % i == 0)
cout<<i;
}

getch();
}

Answer 50

its meant to be int i =1 and not int =1//comeon,,atleast this much Common sense!

Answer 51

u have written like that

Answer 52

I've been getting too many notifications from this. Allow me too answer it.
The answer is: 42

Answer 53

u knw i have nt studied.....looping

Answer 54

must be mis print,,but you should have got it..

Answer 55

u knw i have nt studied.....looping

Answer 56

so what,,you must be knowing int =1 means nothing!

Answer 57

^ No. 42

Answer 58

anyways,,sorry @Compassionate ! :P

Answer 59

42 !

Answer 60

now i got it

Answer 61

hmm,,thanks @Compassionate ! :D

Answer 62

Glad I could help.

Answer 63

SWAGGER TEN MILLION

ONE HUNDRED TRILLION

Answer 64

okay buddy,,its alright,,relax!! ;)

Answer 65

Here I am giving the solution of your problem without using loops.But I am using jump statements to meet your requirements.

Answer 66

thxxx

Answer 67

welcome.

Answer 68

Hello all;
Well, this kind of problem can be solved using two methods : loops or recursive
We start with loop's method because it's the easy one (the loop means do the same thing for many times without rewriting the part of code many times) :

#include<stdlib.h>
#include<iostream>

using namespace std;

int main(){
int n;
cout <<"Enter a number : ";
cin >> n;
if (n<0) cout << "The number must be positive\n";
else {
int facto = 1; //this variable will contain the value of the factorial
for(int i=1; i<=n ;i++){ // Here the loop, it means do this work for n times.
facto = facto * i; //Calculate the factorial.
}
cout <<"The factorial of " <<n <<" is : " <<facto <<"\n";
}
system("pause");
}

The second is the recursive method :

#include<stdlib.h>
#include<iostream>

using namespace std;
//This is a recursive function that can calculate the fact.
int fact(int n){ //We give it the number n as an argument.
if(n==0 || n==1) return 1; //if n =1 or 0 then the value of factorial is 1
else return n*fact(n-1); // if n>1 then call again the function fact and so on.
}

int main(){
int n;
cout <<"Enter a number :"; // simple message
cin >> n; //read the value of n
if (n<0) cout << "The number must be positive\n";
else cout <<"The factorial of " <<n <<"is : " <<fact(n) <<"\n";
system("pause");
}

I hope this help you a little, good luck.

Answer 69

I think the problem needs to be clarified. Factors or Factorial? This two is not the same thing. I said this because some of you give Factors solution, and the other give Factorials solution.