p*****2
发帖数: 21240 | 1 下边这道题，我优化了半天也不行。有标准求common divisor的快速算法吗？
By 2312 there were n Large Hadron Colliders in the inhabited part of the
universe. Each of them corresponded to a single natural number from 1 to n.
However, scientists did not know what activating several colliders
simultaneously could cause, so the colliders were deactivated.
In 2312 there was a startling discovery: a collider's activity is safe if
and only if all numbers of activated colliders are pairwise relatively prime
to each other (two numbers are relatively prime if their greatest common
divisor equals 1)! If two colliders with relatively nonprime numbers are
activated, it will cause a global collapse.
Upon learning this, physicists rushed to turn the colliders on and off and
carry out all sorts of experiments. To make sure than the scientists'
quickness doesn't end with big trouble, the Large Hadron Colliders' Large
Remote Control was created. You are commissioned to write the software for
the remote (well, you do not expect anybody to operate it manually, do you?).
Initially, all colliders are deactivated. Your program receives multiple
requests of the form "activate/deactivate the i-th collider". The program
should handle requests in the order of receiving them. The program should
print the processed results in the format described below.
To the request of "+ i" (that is, to activate the i-th collider), the
program should print exactly one of the following responses:
"Success" if the activation was successful.
"Already on", if the i-th collider was already activated before the request.
"Conflict with j", if there is a conflict with the j-th collider (that is,
the j-th collider is on, and numbers i and j are not relatively prime). In
this case, the i-th collider shouldn't be activated. If a conflict occurs
with several colliders simultaneously, you should print the number of any of
them.
The request of "- i" (that is, to deactivate the i-th collider), should
receive one of the following responses from the program:
"Success", if the deactivation was successful.
"Already off", if the i-th collider was already deactivated before the
request.
You don't need to print quotes in the output of the responses to the
requests.
Input
The first line contains two space-separated integers n and m (1 ≤&#
8201;n, m ≤ 105) — the number of colliders and the
number of requests, correspondingly.
Next m lines contain numbers of requests, one per line, in the form of
either "+ i" (without the quotes) — activate the i-th collider, or "- i" (
without the quotes) — deactivate the i-th collider (1 ≤ i&#
8201;≤ n).
Output
Print m lines — the results of executing requests in the above given format
. The requests should be processed in the order, in which they are given in
the input. Don't forget that the responses to the requests should be printed
without quotes.
Sample test(s)
input
10 10
+ 6
+ 10
+ 5
- 10
- 5
- 6
+ 10
+ 3
+ 6
+ 3
output
Success
Conflict with 6
Success
Already off
Success
Success
Success
Success
Conflict with 10
Already on | l*****a
发帖数: 14598 | 2 too long...
.
prime
【在 p*****2 的大作中提到】
: 下边这道题，我优化了半天也不行。有标准求common divisor的快速算法吗？
: By 2312 there were n Large Hadron Colliders in the inhabited part of the
: universe. Each of them corresponded to a single natural number from 1 to n.
: However, scientists did not know what activating several colliders
: simultaneously could cause, so the colliders were deactivated.
: In 2312 there was a startling discovery: a collider's activity is safe if
: and only if all numbers of activated colliders are pairwise relatively prime
: to each other (two numbers are relatively prime if their greatest common
: divisor equals 1)! If two colliders with relatively nonprime numbers are
: activated, it will cause a global collapse.
| p*****2
发帖数: 21240 | 3
简单的话就是求两个数的common divisor.
【在 l*****a 的大作中提到】
: too long...
:
: .
: prime
| H***e
发帖数: 476 | 4 greatest common dividor?
用 euclid吧，很快的
【在 p*****2 的大作中提到】
:
: 简单的话就是求两个数的common divisor.
| x******a
发帖数: 6336 | 5 先求gcd,然后求gcd的所有divisor?
【在 p*****2 的大作中提到】
:
: 简单的话就是求两个数的common divisor.
| p*****2
发帖数: 21240 | 6
这题只要求common divisor 或 1 或 非1就可以了。有什么快的吗？
【在 H***e 的大作中提到】
: greatest common dividor?
: 用 euclid吧，很快的
| x******a
发帖数: 6336 | 7 1肯定是吧
直接return true？
【在 p*****2 的大作中提到】
:
: 这题只要求common divisor 或 1 或 非1就可以了。有什么快的吗？
| p*****2
发帖数: 21240 | 8
不好意思，是只有1,或者有其他的。
【在 x******a 的大作中提到】
: 1肯定是吧
: 直接return true？
| x******a
发帖数: 6336 | 9 Binary GCD algorithm
?
【在 p*****2 的大作中提到】
:
: 不好意思，是只有1,或者有其他的。
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