ALICESIE - Alice Sieve

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Alice has recently learned to use the Sieve of Eratosthenes, an ancient algorithm for finding all prime numbers up to any given limit. As expected, she was really impressed by it's simplicity and elegancy.

Now, she has decided to design her own sieve method: The Sieve of Alice, formally defined by the following procedure, which determines the Sieve of Alice up to a given limit N.

  1. Create a list of consecutive integers from N to 2 (N, N-1, N-2, ..., 3, 2). All of those N-1numbers are initially unmarked.
  2. Initially, let P equal N, and leave this number unmarked.
  3. Mark all the proper divisors of P (i.e. P remains unmarked).
  4. Find the largest unmarked number from 2 to P – 1, and now let P equal this number.
  5. If there were no more unmarked numbers in the list, stop. Otherwise, repeat from step 3.

Unfortunately, Alice has not found an useful application for it's Sieve. But she still wants to know, for a given limit N, how many integers will remain unmarked.


The first line contains an integer T, the number of test cases (1 ≤ T ≤ 10^4) . Each of the next T lines contains an integer N (2 ≤ N ≤ 10^6).


Output T lines, one for each test case, containing the required answer.




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jawad_cs: 2016-12-01 10:06:30

LOL...i was fooled..hehe

aman224: 2016-11-25 21:22:18

too easy....

davidgalehouse: 2016-10-17 06:03:33

Surely if Alice is designing her own sieves then she would be able to answer this question herself.

suvro_datta: 2016-09-26 21:52:00

( num + 1 ) /2

sri: 2016-09-25 13:15:18

three if's that's it!! Simple but portrayed as hard!!!

hamjosh1: 2016-09-15 14:18:32

pen paper :P

smtcoder: 2016-09-03 22:16:45

HAHAHAHA...this question is a big joke..but will make you feel like a joker if you just start making (sieve like) codes for this question without analysing it for sometime..
DONT TAKE HINTS FOR THIS QUESTION..lets see yourself whether this question fools you or not..

iit2015504: 2016-08-07 19:34:29


sarthak_8: 2016-07-30 23:26:10

Better to move on to another problem as there's nothing to learn here, do it only if you want points. Observation will get you an AC.

try2catch: 2016-02-02 10:20:01

Accepted in 1st attempt. Feeling proud. I dedicate my success to my parents,teachers and friends. I worked hard for two months to solve these kind of problems. Never dreamt better than this.
Now I'm ready for next ACM ICPC. ^_-

Last edit: 2016-02-02 10:20:36

Added by:Paulo Costa
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64