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Description: 北京大学ACM比赛题目
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Every even number greater than 4 can be
written as the sum of two odd prime numbers.
For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach s conjecture for all even numbers less than a million.
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Author: pengfam |
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Description: 北京大学ACM比赛题目
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Every even number greater than 4 can be
written as the sum of two odd prime numbers.
For example:
8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.
Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach s conjecture for all even numbers less than a million.-Peking University ACM Competition Title In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: Every even number greater than 4 can be written as the sum of two odd prime numbers. For example : 8 = 3+ 5. Both 3 and 5 are odd prime numbers. 20 = 3+ 17 = 7+ 13. 42 = 5+ 37 = 11+ 31 = 13+ 29 = 19+ 23 . Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.) Anyway, your task is now to verify Goldbach s conjecture for all even numbers less than a million.
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Author: pengfam |
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Description: 北大在线ACM系统中的经典解题方法思路还有源码。其中是按ACM中的题号作为标题的。-ACM System sharPKUngfu online classic problem-solving methods also source ideas. Which is based on ACM
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Author: jackie |
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Description: ACM源码集合(不是音频编码的ACM), 里面有些很经典的算法代码, 参考ACM竞赛的,有很大的参考价值,需要的可以下载参考-ACM-source collection (not the audio coding of ACM), which some of the classic algorithm code, reference ACM contest, have a great reference value, can download the necessary reference
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Author: yanxiaobin |
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Description: Usage: yi=akimai(x,y,xi)
Given vectors x and y (of the same length)
and the array xi at which to interpolate,
fits piecewise cubic polynomials and returns
the interpolated values yi at xi.
Ref. : Hiroshi Akima, Journal of the ACM, Vol. 17, No. 4, October 1970,
pages 589-602.
Programmer: N. Shamsundar, University of Houston, 6/2002
Correction to lines 32-33, 9/2004, motivated by Gilford Ward,
to make routine work correctly for linear data.
Notes: Use only for precise data, as the fitted curve passes through the
given points exactly. This routine is useful for plotting a pleasingly
smooth curve through a few given points for purposes of plotting.-Usage: yi=akimai(x,y,xi)
Given vectors x and y (of the same length)
and the array xi at which to interpolate,
fits piecewise cubic polynomials and returns
the interpolated values yi at xi.
Ref. : Hiroshi Akima, Journal of the ACM, Vol. 17, No. 4, October 1970,
pages 589-602.
Programmer: N. Shamsundar, University of Houston, 6/2002
Correction to lines 32-33, 9/2004, motivated by Gilford Ward,
to make routine work correctly for linear data.
Notes: Use only for precise data, as the fitted curve passes through the
given points exactly. This routine is useful for plotting a pleasingly
smooth curve through a few given points for purposes of plotting.
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Author: rz |
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Description: acm.timus 1726. Visits
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Author: ahaha |
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Description: 7月17日是Mr.W的生日，ACM-THU为此要制作一个体积为N*pi的M层生日蛋糕，每层都是一个圆柱体。
设从下往上数第i(1 <= i <= M)层蛋糕是半径为Ri, 高度为Hi的圆柱。当i < M时，要求Ri > Ri+1且Hi > Hi+1。
由于要在蛋糕上抹奶油，为尽可能节约经费，我们希望蛋糕外表面（最下一层的下底面除外）的面积Q最小。
令Q = S*pi
请编程对给出的N和M，找出蛋糕的制作方案（适当的Ri和Hi的值），使S最小-July 17 is the birthday of Mr.W, ACM-THU do this want to make a volume of N* pi M-layer birthday cake, each layer is a cylinder.
Set number the bottom section i (1 <= i <= M) layer cake with a radius Ri, a cylindrical height of Hi. When i <M, the requirement Ri> Ri+ 1 and Hi> Hi+ 1.
Due to the cake rubbed cream, as much as possible to save funds, we hope the outer surface of the cake (the lowest layer of the underside of the exception) Q smallest area.
So that Q = S* pi
Please programming given N and M, find the cake production program (the appropriate value of Ri and Hi), the S minimum
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Author: wxs |
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