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在 ACM/ICPC 地区赛中，参赛队可以直接提交题目答案，但若答案错误，则再次提交时最后的成绩会受到影响提交的次。比赛的测试系统对每次提交的程序进行评判，结果是 AC 或者某种错误，参赛队能看到这个结果。
为了奖励优秀的队伍同时确定进军世界总决赛的资格名单，当两个队伍做出的题目数量相同时，会按照其使用的时间来进行进一步的排名。时间有两个部分，第一是总的解题时间，二是惩罚时间。所谓惩罚时间是指提交程序未通过时被罚的时间，每一次未通过的提交，都会在最终用于排名的时间中增加 20 分钟。对于没解决的题目不计时。
程序将读入一张运行结果清单，然后打印出前三名的成绩。
-the ACM / Illinois regional competition, participating teams can answer directly to the topic, but the answer wrong. again at the time of submission of the final results will be affected the times. Competition for each test system of evaluation procedures, the result is AC or some mistakes, teams can see the result. To reward outstanding contingent also set to enter the final of the world's total eligible list, When the two teams made the subject of the number of the same, in accordance with their use of time to carry out further in the rankings. Time has two parts, the first is the general problem solving time is two time penalties. The so-called punishment period of the submission process did not pass the time at halftime, and every time the author did not pass, will the rankings for
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Author: 张日天 |
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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: ACM题目您的文件包然后写出其具体功能(至少要20个字)。不要让站长把时间都花费
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Size: 884 |
Author: dcfeng |
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Description: Shanghai Jiao Tong University ACM-ICPC Team Selection Contest
Stage I
May 17th, 2006 18:00 » 20:30
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Size: 79022 |
Author: darren |
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Description: 这是ACM的一道试题.内容是:“蛇和梯子”是一个在N*N（0<N<=20)的方格棋盘上进行的游戏.-This is a test of the ACM. It reads : "snakes and ladders" is one of the N * N (0lt; Nlt; = 20) in the box on the chessboard of the game.
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Author: 小丽儿 |
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Description: 这是ACM的一道试题.内容是:“蛇和梯子”是一个在N*N（0<N<=20)的方格棋盘上进行的游戏.-This is a test of the ACM. It reads : "snakes and ladders" is one of the N* N (0lt; Nlt; = 20) in the box on the chessboard of the game.
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Size: 35840 |
Author: 小丽儿 |
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Description:
在 ACM/ICPC 地区赛中，参赛队可以直接提交题目答案，但若答案错误，则再次提交时最后的成绩会受到影响提交的次。比赛的测试系统对每次提交的程序进行评判，结果是 AC 或者某种错误，参赛队能看到这个结果。
为了奖励优秀的队伍同时确定进军世界总决赛的资格名单，当两个队伍做出的题目数量相同时，会按照其使用的时间来进行进一步的排名。时间有两个部分，第一是总的解题时间，二是惩罚时间。所谓惩罚时间是指提交程序未通过时被罚的时间，每一次未通过的提交，都会在最终用于排名的时间中增加 20 分钟。对于没解决的题目不计时。
程序将读入一张运行结果清单，然后打印出前三名的成绩。
-the ACM/Illinois regional competition, participating teams can answer directly to the topic, but the answer wrong. again at the time of submission of the final results will be affected the times. Competition for each test system of evaluation procedures, the result is AC or some mistakes, teams can see the result. To reward outstanding contingent also set to enter the final of the world's total eligible list, When the two teams made the subject of the number of the same, in accordance with their use of time to carry out further in the rankings. Time has two parts, the first is the general problem solving time is two time penalties. The so-called punishment period of the submission process did not pass the time at halftime, and every time the author did not pass, will the rankings for
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Author: 张日天 |
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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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Size: 2048 |
Author: pengfam |
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Description: Shanghai Jiao Tong University ACM-ICPC Team Selection Contest
Stage I
May 17th, 2006 18:00 » 20:30
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Size: 78848 |
Author: darren |
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Description: poj题目，北京大学acm队选拔用题，经典好题-POJ topic acm Peking University team selection by title, classic good title
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Size: 513024 |
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Description: 北京大学的一些acm代码。还有一些资料。-Acm code of some of Peking University. There are some information.
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Size: 138240 |
Author: 马聪 |
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Description: acm算法设计：
描述：
国际象棋车可以水平或垂直的任何方在同一行或同一列的棋盘。
找到一些最短路径的一个车可以从一个角落里一个棋盘的斜对面角落。
输入：
一整数n为行数和列的棋盘。0 < n <=16
输出：
数量的最短路径。
例输入：
3或4
例输出：
6或20 -Description：
A chess rook can move horizontally or vertically to any square in the same row or in the same column of a chessboard.
Find the number of shortest paths by which a rook can move from one corner of a chessboard to the diagonally opposite corner。
Input：
a interger number n is row and column of chessboard. 0 < n <=16
Output：
the number of shortest paths.
Sample Input：
3or4
Sample Output：
6or20
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Size: 211968 |
Author: 张波 |
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Description: Solution archive of ACM ICPC NEERC quaterfinal 2009-2010
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Size: 307200 |
Author: matkarimov |
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Description: ACM POJ 2247.cpp 2051.cpp 1656.c 1326.cpp 1016.cpp保证正确-ACM POJ 2247.cpp 2051.cpp 1656.c 1326.cpp 1016.cpp
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Size: 2048 |
Author: 刘亚 |
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Description: 杭电acm解题报告 详细解析2000-2099 适合acm初学者-acm of HDU
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Size: 825344 |
Author: 尉栋超 |
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Description: 22. 切木头
成绩: 10 / 折扣: 0.9
背景
人们需要把一跟很长的木头切成几段，有一家名为 Analog Cutting Machinery (ACM) 的公司正在经营这一业务。他们根据切割前木头的长度来收费，木头越长、收费越高，并且每切割一次就收一次费。
显而易见，在这里切割木头时，不同的切割顺序就会产生不同的价钱。譬如一跟 10 米长的木头，需要在 2、4、7 米处切开。如果顺序在这三个位置切割，需要的费用是 10 8 6 = 24，因为木头原始长度为 10 米，切掉两米剩 8 米，在四米处切掉剩 6 米。如果按照 4、2、7 的顺序来切割，花费就是 10 4 6 = 20。
任务
你的老板有很多木材要切割，现在他希望你能够帮他找到最便宜的切割方式。
输入
一次输入可能包含多组数据。每一组数据的第一行是木材的长度L (L<=1000)，如果为 0 则表示输入结束。每组数据的第二行是要切割的次数 N (N<=50)，第三行则是切割的位置Ci (0<Ci<L)。以上数据均为整数。
输出
针对每一组输入，输出切割这段木头的最小费用。
-22. Cutting wood
Results: 10/Discount: 0.9
Background
People need to put a cut into the wood with long paragraphs, there is a company called Analog Cutting Machinery (ACM) of the company is operating the business. Them according to the length of wood before cutting charges, the longer the wood, the higher the fees, and each time to receive a fee cut.
Obviously, cutting wood in here, different order will produce different cutting price. For instance, one with 10 meters of wood, you need to cut the 2,4,7 meters. If the sequence of cutting in these three locations, the cost of required 1086 = 24, because the original length of 10 meters of wood, cut two meters 8 meters left, cut off the remaining four meters 6 meters. If the cut in accordance with the order 4,2,7, cost is 1046 = 20.
Task
Your boss has a lot of wood to be cut, and now he wants you to help him find the cheapest way of cutting.
Input
One input may contain multiple sets of data. The first line of each set of dat
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Author: a123 |
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Description: ACM详细解题过程 2000-2099题-2000-2099 ACM detailed process of problem solving questions
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Size: 825344 |
Author: 林舒 |
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Description: This C++ codes are developed by Shin Yoshizawa at the MPII, Saarbruecken, Germany. The method is described in my paper "Free-form Skeleton-driven Mesh Deformations", Shin Yoshizawa, Alexander G. Belyaev, and Hans-Peter Seidel, ACM Solid Modeling 2003, pp. 247-253, June 16-20, 2003, University of Washington, Seattle.
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Size: 10240 |
Author: Niu Wenjie |
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Description: 很好很强大的acm专用书很强大，文件不能说明太短，最少20字-great，and usuful，if you want to be the acmer ，you must try this。
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Size: 10240 |
Author: Timson |
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Description: 访问控制模型研究进展及发展趋势，文件说明太短，至少要20个字-Research status and development trends of Access control model
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Size: 673792 |
Author: wolala |
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