這兩天都是水題，沒有什么想法而言，就是為了熟悉一下感覺。我又給互聯網上增加了不少垃圾日志。
昨晚出了點突發事件，原定的寫程序事件被用來處理事情了。不過后來我倒是有點想明白了，題目只是很少一點時間在看題目，很忙的時候可以先花幾分鐘把題目弄明白了，然后在去忙別的事情，這種件的空隙還能想一想思路。不知POJ有什么辦法可以把題目批量導出的，我就不用天天費勁上網讀題目了，在單位也能抽空瞅兩眼。
我堅持兩件事情：1。決不直接貼題目的代碼 2。每天都寫題目
POJ 3006
計劃打個表，然后直接搞，看了一下數據很小，應該沒有什么問題。
在這里順便題一下素數篩選法。
以下內容來自Wikipedia:
To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:
- Create a list of consecutive integers from two to n: (2, 3, 4, ..., n).
- Initially, let p equal 2, the first prime number.
- Strike from the list all multiples of p less than or equal to n. (2p, 3p, 4p, etc.)
- Find the first number remaining on the list after p (this number is the next prime); replace p with this number.
- Repeat steps 3 and 4 until p2 is greater than n.
- All the remaining numbers in the list are prime
后來，歐拉同學提出了另外一個辦法，讓每個數只要標記一次就ok
A) Start with all the natural numbers except '1' which is not a prime:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ...
^
B) The leftmost number is prime. Multiply each number in the list by this prime and
then discard the products:
(4 6 8 10 12 14 16 18 20 22 24 26 28 30 ... )
These are removed:
4 6 8 10 12 14 16 18 20 22 24 26 28 30
These are left:
2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 ...
^
C) The number after the previous prime is also a prime. Multiply by it each number
in the new list starting from this prime and then discard the products:
(9 15 21 27 33 39 45 51 57 63 69 75 81 87 ...)
These are removed:
9 15 21 27
These are left:
2 3 5 7 11 13 17 19 23 25 29 ...
^
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2010-7-13 23:47
拆掉了自己的惠普V3159筆記本，心情一片大好，竟然裝上以后運行正常了。出來混，遲早要還，之前一直對于拆機器的事情不感興趣，現在可真是被逼得。
我終于可以去中關村賺取那90塊的拆機費了: )
PS:今天中午修好了屋子里面的抽水馬桶，心情一片大好。