The problem is like this - You have given an array A having N integers. Let say G is the product of all elements of A. You have to find the number of distinct prime divisors of G. Example - Input : A = 1, 2, 3, 4 Output : 2
Explanation : Here g = 1*2*3*4 and distinct prime divisor of g are 2 and 3 To total count of distinct prime divisor = 2
Below is the code i wrote but the output which i am getting is wrong -
class prime:
# @param A : list of integers
# @return an integer
def prime(self,num):
if(num>1):
for i in range(2,num):
if(num%i==0):
break
else:
return num
def solve(self, A):
prod = 1
tot = 0
for i in range(0,len(A)):
prod = prod*A[i]
for i in range(0,len(A)):
if(self.prime(A[i])):
if(prod%self.prime(A[i])==0):
tot = tot+1
return tot
A = [1,2,3,4]
prime().solve(A))
class prime:
# @param A : list of integers
# @return an integer
def prime(self,num):
if num == 2: # changes
return num # changes
if(num>2): # changes
for i in range(2,num):
if(num%i==0):
break
else:
return num
def solve(self, A):
prod = 1
tot = 0
for i in range(0,len(A)):
prod = prod*A[i]
for i in range(0,len(A)):
if(self.prime(A[i])):
if(prod%self.prime(A[i])==0):
tot = tot+1
return tot
A = [1,2,3,4]
print(prime().solve(A))
Lines that were changed were commented with # changes
from math import sqrt
from itertools import count, islice
class Prime:
def prime(self, n):
if n < 2:
return False
for number in islice(count(2), int(sqrt(n) - 1)):
if n % number == 0:
return False
return True
def solve(self, A):
prod = 1
tot = 0
for i in range(0, len(A)):
prod = prod * A[i]
if(prod<2):
return 0
if(prod == 2 or prod == 3):
return 1
for i in range(2, prod/2+1):
if(self.prime(i) and prod % i ==0):
tot =tot+1
return tot
A = [1,2,3,4]
print(Prime().solve(A))
After going through the give inputs and outputs by the OP i understood that OP wants to count the number of primes which can completely divide the prod (product of element) and give remainder as 0. Input 1 by OP - g = 1*2*3*4 and distinct prime divisor of g are 2 and 3. Total count of distinct prime divisor = 2 Input 2 by OP - g = 96*98*5*41*80 and distinct prime divisor of g are 2,3,5,7 and 41. Total count of distinct prime divisor = 5
Code for the above problem -
class Solution:
# @param A : list of integers
# @return an integer
def prime(self,num):
if(num==1):
return 0
for i in range(2,(num//2+1)):
if(num%i==0):
return 0
return num
def solve(self, A):
prod = 1
tot = 0
for i in range(0,len(A)):
prod = prod*A[i]
for i in range(1,(prod//2+1)):
pr = self.prime(i)
if(pr):
#77145600
print("Prime :",pr)
if(prod%pr==0):
tot = tot+1
return tot
A = [96,98,5,41,80]
print(Solution().solve(A))
But for this code, the response time is very high. For this input - 96,98,5,41,80 the response time was more than 5 hours. Can anyone provide a better solution for it?
I found a better solution then the above mentioned by me -
Updated new code -
# Python Program to find the prime number
def prime(num):
if(num==1):
return 0
for i in range(2,(num//2+1)):
if(num%i==0):
return 0
return num
# Python Program to find the factors of a number
def findFactors(x):
# This function takes a number and finds the factors
total = 0
for i in range(1, x + 1):
if x % i == 0:
if(prime(i)!=0):
print("Prime : ",prime(i))
total+=1
print("Total : ",total)
return total
# change this value for a different result.
num = 77145600
findFactors(num)
The findFactors function first finds the factors of given number and then by using prime function I am checking whether the found factor is prime or not. If it is a prime number then I am incrementing the total by 1. Execution time is 45 seconds on my system.
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