Showing wrong answer

#1

My code is showing correct answer for 1 test case and wrong answer on remaining cases. Please help

#2
#3

Can you please explain the logic you used , your code is tough to comprehend.

#4

First I found 10^6 prime numbers using prime seive because we have to multiply them by their powers in given array to find n. Then for finding power of different prime no. I use modular exponentiation as power can be large then I multiply them all like ((b[0]^prime[0])(b[1]^prime[1])…so on)=n. Now P=n^((No. of divisors of n)/2) and in my soln No.of divisors of n are po and P can be find by using modular exponentiation as it’s power can be large. Now after this I factorize P and find it’s prime factors with their powers and then for divisors of P, I multiply power of prime factor+1 for all prime factors and take their modulo and print that as answer.

#5

Okay so the question has nothing to do with finding prime numbers, infact to find 10^6 prime numbers you would have to run sieve for a much bigger n.
See this code https://ide.codingblocks.com/s/257746

I am going to break down the code here.
This portion calculates the number of divisors of given number
long long nOfDivisors = 1;

 for (long long i = 0; i<x; i++)
 {
      cin >> vec[i];
      nOfDivisors = (nOfDivisors*(vec[i] + 1)) % p;
 }

The product of all the divisors of a number N is N^(x/2) where x is the number of divisors of N. The number of divisors is given by product of prime powers+1 for N.

Now the next part is basic P&C.

 for (long long i = 0; i<x; i++)
 {
      vec[i] = (nOfDivisors*vec[i])%p;
      vec[i] = (vec[i]*500000004)%p; // 500000004 is mod_inv(2,1000000007);
      ans = (ans*(vec[i] + 1)) % p;
 }
#6

Thanks for your explanation, I got it.

#7

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