[競賽] Al Zimmermann's Programming Contests

現在請你找出 n 個正整數, 然後挑他們其中幾個, 用加減運算符號構成一個運算式, 看所有可能組合中, 你得出的答案可以有幾個不同的質數。
理論上會有 3ⁿ-1 個(你可以把它想成"現在請你找出 n 個正整數, 然後他們各自任意乘上1,0,或-1後再全部加起來, 你得出的答案可以有幾個不同的質數。", 就了解為什麼會有這麼多了)

當 n=3 時, 你會找到幾個??
當 n=4 時, 你會找到幾個??
...
當 n=14 時, 你會找到幾個??

當然找到越多越好....

接下來要談最傷感情的事了 - 怎麼計分??
基本上公式是這樣的

R = the current record for the given N
S = your result for the given N
Q = log₁₀(1-(S/R))
Score = Q/(Q-1)

如果當 n =3 時, 你用了(2,3,5)這組數字, 數字組合搭配, 找到了 4 個不同的質數(S=4)。
2,3,5,7(2+5)
結果可是全部的人中, 居然有人找到了 5個(R=5)。
所以 Q = log₁₀(1-(4/5)) = -0.69897
你在 n=3 的分數就是 Score = -0.69897/(-0.69897-1) = 0.411408

當然, 如果你在某個 n 值中找到了所有人中最多的不同質數, 那麼你的 Score 就直接是 1

最後, 將你在 n=3, n=4...n=14 所有得到的分數加起來, 就是你最後的成績。

以下是範例:

Let us assume that the best submission for n=4 is "8, 41, 52, 78" producing 16 distinct primes, submitted by participant B.
Participant A makes his first submission for the category n=4.
He submits the 4 numbers "18, 23, 27, 43" thus producing 10 distinct primes among the 80 possible sums.
Then A's score for n=4 is computed by Q=log₁₀(1-10/16)=-0.425969, Score=Q/(Q-1)=-0.425969/(-0.425969-1)=0.2987
Later participant C submits "24,30,36,77" producing a new record with 17 distinct primes.
C gets a score of 1.0 for n=4.
A's score for n=4 drops from 0.2987 to 0.2782, i.e. A will see a reduction of his total score by 0.0196.
The total score of the previous record holder B will drop significantly by 0.4483, because the updated score for his 16 primes submission becomes 0.5517, computed by Q=log₁₀(1-16/17)=-1.230449, Score=-1.230449/-2.230449=0.5517.

(沒力了, 自己去翻譯, 我還有好幾篇文章沒打咧)

當然, 還有一些注意事項:

You may submit as many solutions as you want, for any n within the range 3<=n<=14.
The number of submissions has no influence on the score.
The sum of all the numbers must not exceed 2³²-1.
Submissions containing less than 3 or more than 14 numbers are rejected by the scorer.
The entries of a solution may be submited in any order, i.e. "10 20 31 50" or "50 10 31 20" are both valid and equivalent submissions for n=4.
The numbers in the submission have to be separated by at least one non-digit character.
Any non-digit character will be converted into a space.

When you submit a solution that produces less prime sums than a previous solution you had submitted before for the same n, your score remains unchanged.
If more than one team has the same total score based on the score sum computed with high precision at the end of the contest, the ranking for this score will be done based on the time when the affected teams submitted the last solution improving their score (earlier is better).
The ranking is always based on high precision scores, although scores are only displayed with 4 decimal digits.

Whenever a participant submits a solution that is better than the current best solution for a given n, the corresponding scores of all participants are recomputed and adjusted immediately.

(....原因同上)

接下來有個最重要的事: 那....我要自己一個一個去算嗎??
n=3, n=4 時還好, 要是 n=14 我不是要算到死??

別擔心, 你可以到這個網址 http://www.recmath.org/contest/test.php 來測試這組數字可以找出多少個不同的質數。
所以就放心大膽地做吧....加油~~

----

最後, 要來公布最重要的事了....
這期的獎金數目....

咳, 為了要彰顯這次競賽的價值。
所以....
第一名我們決定不頒獎金, 改頒知名藝術創作家 Bathsheba Grossman 的作品其中一件。
http://www.bathsheba.com/sculpt
http://www.bathsheba.com/math

第二名, 也可以獲得同一藝術家的作品 .... 迷你版。
http://www.bathsheba.com/sculpt/minimetal.html
http://www.bathsheba.com/math/minimath.html

為了讓您感受到這些藝術作品的價值, 所以小弟特別去網路上找了這位 BG 女士的介紹:
http://intozgc.zhuaxia.com/item/842030868
(其實老實說, 作品真的挺不錯, 有空間幾何的思維)

謝謝您忠實地閱讀完本文...
也請忘了那 270 美元, 它並不在本次比賽存在
如果您有郭台銘的身價, 請省下您買 Wii 的預算, 掏出您的荷包....中的信用卡, 贊助 Al Zimmermann's Programming Contests 這個網站的運作(PayPal)

再次感謝您看完這篇賺不了錢還可能要花錢的比賽文章....
祝您有個好夢....

[競賽] Al Zimmermann's Programming Contests

祈求一顆天真的心

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