# NUMFACT - Editorial

Author: Vamsi Kavala
Tester: Roman Rubanenko
Editorialist: Bruno Oliveira

Cakewalk

# PRE-REQUISITES

Simple Math, Integer Factorization

## Problem:

You are given a very large number represented as a product of N numbers.
Given this number representation, you need to find the number of distinct factors of the original number which is formed by the product of given N numbers.

## Quick Explanation:

We can factorize each one of the N given numbers into its prime factors. Then we find the number of occurrences of each prime factor, say they are a1, a2,…aK, if we have K distinct prime factors. Our answer is simply: (a1+1)(a2+1)(…)*(aK+1).

## Detailed Explanation:

This problem relies on some knowledge of divisor function. Divisor functions returns the number of positive and distinct divisors of a number. Let’s call it d(x).

• Some properties of the divisor function:

We now look into some important properties of the divisor function:

For a prime number p, we have d§ = 2, as there are only two numbers which divide a prime number:1 and itself.

Now, it’s a known fact that this function is multiplicative but not completely multiplicative. This means that if two numbers, say, a and b are there such that gcd(a, b) = 1, then the following holds:
d(a*b) = d(a)*d(b).

This allows us to deduce the important relationship, that is the key of solving this problem:

For a prime number, p, we have: d(p^n) = n+1.

Now, it’s easy to understand that all we need to do is to factorize all the N given numbers into its prime factors, and, for each prime factor we also need to count how many times it appears (that is, we need to know the exponent of each prime factor).

Once we have this count with us (which can be done using integer factorization and for example, the set and map data structures, one to guarantee uniqueness of the factors and the other to save the number of occurences for each unique prime factor), all we need to do is to multiply all these numbers plus one together and we will obtain our answer.

As an example, consider the number:

504 = 2^3 * 3^2 * 7^1

The number of distinct divisors of 504 is then (3+1) * (2+1) * (1+1) = 24.

Applying this process to all numbers yields the answer to the problem

### SETTER’S SOLUTION

Can be found here.

### TESTER’S SOLUTION

Tester’s solution will be uploaded soon.

6 Likes

Can you tell me where my code fails… I have applied same concept i think…
http://www.codechef.com/viewsolution/2304397

it not works for prime numbers above 1000 like 1009,1013 etc

it works… I have used “other” array for those primes
Can u provide any testcase where it fails…

Either you can use a sieve to extend your primes array or just put more primes in the primes array.

here’s a sample code for sieve.
http://ideone.com/arhRSv

@shasha1s2 wen u update ‘other’ array to count prime u do not account for the fact that the prime number being counted may be the same…in other words, the value of other[i] (according to your code) will never be more than 1…going by that the test case
5
999983 999983 999983 999983 999983
will give 32 as per your code, while ans is 6.

2 Likes

thanx @amitrc17 got my error…

It gives a TLE when i use the function ‘prime’ to find the prime factorisation of the number.Can someone please tell me where i went wrong? Thanks in advance.
Here is the link to my code : http://ideone.com/kgzICq

@roshi…it will most certainly give a TLE as the time complexity is very high!!!

try using Sieve of Eratosthenes…see the time difference…Naive approachseive!!!

1 Like

What is wrong in my code??

It gives correct answers to all the numbers with which I have checked.

@shubham26
if you look at the constraints .They are :-
N<=10
Ai<=1000000

and as in your code you multiply each Ai to get a number which is product of all of them.
Suppose if all the numbers are 10^6 and N=10 and they are multiplied N times,the number becomes (10^6)^10=10^60 which is far beyond the range of long(range of the order 10^9)( (even its far away from long long(range of the order 10^18)).

1 Like

@dhruvagga : thanks…I got it.

In the setter’s solution what is the use of this code fragment
if(a[i]!=1)
{
if(m.find(a[i])!=m.end())
m[a[i]]++;
else
m.insert(MP(a[i],1));

from line 85 to 90
I put that in comments and ran the program and it gave the correct o/p for test cases. I don’t understand it’s use. After dividing with all the prime numbers shouldn’t a[i] necessarily be 1.

1 Like

#include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
if(t>=1 && t<=100)
{
while(t–)
{
int n;
scanf("%d",&n);
if(n>=1&&n<=10)
{
long int a, total=1;
while(n–)
{
scanf("%ld",&a);
if(a>=2&&a<=1000000)
total*=a;
}
int count=0;
for(int i=1;i<=total;i++)
{
if(total%i==0)
count++;
}
printf("%d\n",count);
}
}
}
return 0;
}

### why time limit exist???

Can anyone provide critical input. My solutions keeps getting WA, however, when comparing to AC code gives same output.

from sys import stdin
from collections import Counter
from math import sqrt,ceil

def primes(n):
""" Input n>=6, Returns a list of primes, 2 <= p < n """
correction = (n%6>1)
n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[      ((k*k)/3)      ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]

def factor(n):
upper = n
i = 0
while sieve[i] <= ceil(sqrt(upper)):
while n%sieve[i]==0:
n/=sieve[i]
if sieve[i] not in factors:
factors[sieve[i]]=0
factors[sieve[i]]+=1
i+=1
if(n>1):
if upper not in factors:
factors[upper]=0
factors[upper]+=1

sieve = primes(10000)
factors = {}

def solve():
#print([i for i in numbers])
factors.clear()
for num in numbers:
factor(num)
tot = 1
for num in factors:
#print("fac",num,factors[num])
tot *= (factors[num]+1)
print tot

def main():
for i in range(1,tc+1):
solve()

if __name__=="__main__":
main()

In your code, you are checking whether a[i] == 1 towards the end. But can you give me a case where it wont be equal to 1. Wont the prime numbers stored in the list make sure that the smallest number is being factorized first and then proceed to the largest??

A simple modification of Sieve of Erastothenes will also prove to be helpful. The following is what I used.

f[i] == 0 if i is prime, and if i is prime, then f[i] will have the smallest prime that will divide it.

for (int i = 2; i <= N; i++)
if (!f[i]) for (int j = i+i; j <= N; j += i) if (!f[j]) f[j] = i;

f[i] will help a lot to find all the prime divisors of a number.

http://www.codechef.com/viewsolution/2894661

A shorter solution with the same concept + the subtle modification of the sieve to store prime factor info.

whats the problem wid following ? i got a WA

#include<stdio.h>
#include<math.h>
int primes[1000000];
int a[1000000];
void gen_primes(int n)
{
int i,j;
for(i=0;i<n;i++)
primes[i]=1;
for(i=2;i<=(int)sqrt(n);i++)
{
if(primes[i])
{
for(j=i;j*i<n;j++)
{
primes[i*j]=0;
}
}
}
}
main()
{
int t,n,i;
long long int p,x;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<(int)sqrt(n);i++)
{
a[i]=0;
}
gen_primes(n);
while(n--)
{
scanf("%lld",&x);
for(i=2;i<=(int)sqrt(n);i++)
{
if(primes[i])
{
if(x%i==0)
{
a[i]=a[i]+(long long int)((double)log(x)/(double)log(i));
}
}
}
}
p=1;
for(i=2;i<=(int)sqrt(n);i++)
{
if(a[i])
{
p=p*(a[i]+1);
}
}
printf("%lld\n",p);
}
return 0;
}

my code is running fine on the first time… but in second test case giving erronous output… please help…

#include<stdio.h>

int freq[1000000];
int sum;

void count(long long int x)
{
int i;
for(i=2;;i++)
{
while(x%i==0)
{
freq[i]++;
x=x/i;
sum++;
}
if(x==1)
break;
}
}

int main(void)
{
int t,n,p,y,i;
long long int x;
scanf("%d",&t);
while(t--)
{
p=1;
y=0;
sum=0;
scanf("%d",&n);
for(i=0;i<n;i++)
freq[i]=0;
while(n--)
{
scanf("%lld",&x);
count(x);
}
for(i=0;;i++)
{
if(freq[i]!=0)
{
p=p*(freq[i]+1);
y=y+freq[i];
if(y==sum)
break;
}
}
printf("%d\n",p);
}
return 0;
}
//