# 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,ā¦aK, if we have K distinct prime factors. Our answer is simply: (a1+1)(a2+1)(ā¦)*(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. Letā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(p) = 2, as there are only two numbers which divide a prime number:1 and itself.

Now, itā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, itā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

### SETTERāS SOLUTION

Can be found here.

### TESTERāS SOLUTION

Testerās solution will be uploaded soon.

9 Likes

Can you tell me where my code failsā¦ I have applied same concept i thinkā¦
http://www.codechef.com/viewsolution/2304397

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

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

hereās a sample code for sieve.
http://ideone.com/arhRSv

1 Like

@shasha1s2 wen u update āotherā array to count prime u do not account for the fact that the prime number being counted may be the sameā¦in other words, the value of other[i] (according to your code) will never be more than 1ā¦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

It gives a TLE when i use the function āprimeā 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 : kgzICq - Online C++ Compiler & Debugging Tool - Ideone.com

@roshiā¦it will most certainly give a TLE as the time complexity is very high!!!

try using Sieve of Eratosthenesā¦see the time differenceā¦Naive approachā¦seive!!!

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

In the setterā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 donāt understand itās use. After dividing with all the prime numbers shouldnāt a[i] necessarily be 1.

1 Like

#include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
if(t>=1 && t<=100)
{
while(tā)
{
int n;
scanf("%d",&n);
if(n>=1&&n<=10)
{
long int a, total=1;
while(nā)
{
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()``````

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 timeā¦ but in second test case giving erronous outputā¦ please helpā¦

``````#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;
}``````

Where Am i Going Wrong , Its Always SIGSEV and TLE on my Submission :ā( , Plz help me Fast i Need to Brace up For IOI

``````#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void solve(long long int *primes,long long int *n )
{

long long int *z = (long long int *)calloc(1001,sizeof
(long long int));

while(*n>1)
{
for(long long int i=0;i<168;i++)
{
if(*n%primes[i]==0){z[i]++;*n= *n/primes[i];i--;}
}

}
long long int prod=1;
for(long long int k=0;k<168;k++)if(z[k]!=0)prod*=(z[k]+1);
printf("%lld \n",prod);
//for(long int i=0;i<n&&primes[i]!=0;i++)printf("%ld \n",primes[i]);
}
int main()
{
long long int *primes;
primes = (long long int *)calloc(1001,sizeof(long long int));
primes[0]=2;
primes[1]=3;
long long int index=2;
for(long long int z=5;z<=1001;z+=2)
{
long long int fact=0;
for(long long int i=0;i<1001&&primes[i]!=0;i++)
{
if(z%primes[i]==0){fact++;break;};
}
if (fact==0)
{
primes[index]=z;
index++;
}
}
long long int t;
scanf("%lld",&t);
while(t--){
long long int n,prod=1,temp;
scanf("%lld",&n);
for(long long int i=0;i<n;i++){scanf("%lld",&temp);prod*=temp;}

solve(primes,&prod);
}
}``````

Hey whats wrong in my code here?

``````
program FACTORS;

uses Math;
var
t,n,num,product:longint;

function factors(fact:longint):longint;
var
no:longint = 0;
x:longint;
begin
for x := 1 to floor(sqrt(fact)) do
begin
if fact mod x = 0 then
begin
if x*x = fact then inc(no,1)
else inc(no,2);
end;
end;
factors := no;
end;
begin
while t <> 0 do
begin
product := 1;
while n <> 0 do
begin
product := product * num;
dec(n);
end;
writeln(factors(product));
dec(t);
end;
end.

``````

I am getting runtime error NZEC.

I have been trying to debug my code for so long still not able to figure out why it is giving a WA for the basic cases. Please if anyone could tell me where am I going wrong?
This is the link to my solution.
https://www.codechef.com/submit/complete/8669014

nice explanation.
I used sieve of eratosthenes to get all prime factors of number

https://www.codechef.com/viewsolution/14084014
WHY IS THIS SOLUTION GETTING A WA ON LAST CASE PLZ HELP!!

it worksā¦ I have used āotherā array for those primes
Can u provide any testcase where it failsā¦