Topcoder SRM 643 Div1 250<peter_pan>

Topcoder SRM 643 DIV1 250

Problem

Give an integer n, and then give a Vector<long long>v; n can be expressed as a product of several prime numbers, N=P0*P1*P2*......*PN, we assume that the P0,P1,..., PN is monotonic, then v is stored in the subscript is an even number of N of the mass factor p0,p2,p4,..., p (2k). Now asked to write a program that returns a Vector<long long>ans; Ans Stores the p0,p1,p2,..., PN.

Limits

Time Limit (MS): 2000

Memory Limit (MB): 256

N: [2,10^18]

Solution

n is continuously removed from the number of V (N/=v[i];) and a new n is obtained, which is recorded as N1. The N1 is then decomposed factorization. In [1,10^6] scanning, will N1 decomposition, to obtain a new N1, recorded as N2. If the N2 is not 1, it can be proved that N2 must be a prime number, add ans. The ANS is sorted and returned.

More

Worry about the following scenario, N=2*P1,P1 is a prime number with a level of 1e17. So in the decomposition of factorization, you can not use O (n^0.5) algorithm, otherwise it will time out, and used in [1,10^6] scan to decompose N1. The following shows that N2 must be a prime number, considering the mass factor of n , there are up to two qualitative factors PI,PJ will be greater than 10^6 (the other mass factor is less than or equal to 10^6, then in the [1,10^6] scanning to decompose N1 has been found). Assuming there are two qualitative factors pi,pj greater than 10^6, then because the given V is ordered, PI,PJ is the largest of all factorization, so PI,PJ must have and only one appears in V, so n2=pi or pj,n2 is prime; assuming there is only one pi greater than 10^6, then n2= PI,N2 is a prime number.

Complexity

Time Complexity:o (1e6)

Memory Complexity:very Small

Source

Topcoder SRM 643 DIV1 250

Code

Topcoder SRM 643 Div1 from My Github

Topcoder SRM 643 Div1 250<peter_pan>