什么是USACO?USACO晋级规则说明!附USACO课程

参加USACO是对于计算机领域学生来说一个非常有益的选择。除了通过获得证书为进入美国本科提供加分外,还有机会掌握计算机算法技能、获得就业机会、追求创业梦想,并提升自己的脑力和逻辑思维能力。

什么是USACO?

USACO是美国计算机奥林匹克学术活动(美国信息学奥林匹克学术活动)的缩写。它是一个著名的在线题库,专门为信息学学术活动选手准备的。这项学术活动在全球范围内广泛开放,允许全球中小学生参与,且完全免费。参与者只需要拥有想法和野心,就可以参加这个令人激动的比赛。

USACO的学术活动内容主要涉及算法和程序设计两个方面。通过参加USACO学术活动,学生不仅可以培养他们的编程思维和问题解决能力,还可以在他们今后申请大学时获得很大帮助。USACO成绩在常春藤大学的招生过程中被视为重要的参考指标之一,因此参与USACO学术活动可以极大地提高学生的竞争力。可以说,USACO是一项极具含金量的学术活动。

USACO晋级规则如下:

每个USACO学术活动组别共有3道题目,总分为1000分。

代码提交后,系统会自动给出评分。每个问题的分值为333.333分,总分为1000分。

如果取得满分,系统将提示直接晋级。这样可以在本次月赛中挑战更高难度的试题。换句话说,如果你在某个组别的所有题目上都取得满分,你可以跳级到更高级别的学术活动(即下一个组别)。否则,如果你没有取得满分,你将等候分数线的公布。

一般情况下,在每次月赛考试结束后,会公布晋级分数线。若你的总分高于或等于分数线(通常高于750分或800分),则你可以在下一个月的比赛中参加更高级别的学术活动。

培训课程

1.我们的USACO课程内容与USACO考试的要求紧密契合,有助于学生全面掌握所需的算法知识。

2.USACO课程重点突出算法考点知识。算法是USACO考试的核心内容,在培养学生的编程能力和思维能力方面起着至关重要的作用。通过深入学习和练习算法,学生能够有效提升自己在比赛中的表现,并且培养出解决问题的能力。

3.授课教师均来自海内外名校,拥有丰富的学术活动辅导经验,多次带领学生在USACO学术活动中取得突出成绩。

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USACO竞赛考多少分可以晋级?USACO竞赛有何价值?

USACO计算机学术活动是美国的一项面向中学生的信息学学术活动。对于海外留学申请理科名校以及CS专业的同学来说,拿到USACO学术活动的证书对申请非常有帮助。USACO不同于其他奥林匹克学术活动,不仅针对美国学生,也对其他国家的学生非常友好,并且鼓励大家参与。

USACO学术活动考多少分可以晋级?

USACO学术活动每个组别都有三道题,满分1000分,每个问题的分值都是333.333分。从近三年的分数线来看,USACO学术活动的分数线相对稳定,一般高于750或800分的分数通常就能晋级。

USACO学术活动参赛价值

1.参加USACO学术活动是申请美本的一项加分项,且具有高校认可度。以MIT为例,2024届早申录取的两名中国学生中,一位学生在中国的NOI比赛(相当于美国的USACO比赛)中获得金牌(全国前50名),入选信息学国家集训队,并且凭借这一成就成功保送清华大学。这一案例说明,USACO学术活动成绩在高校招生中具有重要的参考价值,特别是在计算机科学相关专业的申请中。

2.参加USACO学术活动可以让学生掌握重要的计算机算法技能,这为他们追求从事互联网领域工作创造了机会。在当今互联网行业快速发展的背景下,计算机算法的掌握成为从事这一领域工作的基础。通过参加USACO学术活动,学生可以不仅熟练掌握各类算法,还能培养解决问题的能力、培养良好的编程习惯和代码管理能力等,这些都是互联网行业工作中必备的技能。掌握了这些技能,学生将有更多机会获得在互联网行业工作并在工作中稳定发展、晋升道路明显的机会。

3.参加USACO学术活动还能够提高学生的脑力和逻辑能力。学术活动涉及的问题往往需要学生进行深入思考和分析,培养了他们的逻辑思维能力和问题解决能力。通过不断解决各类算法问题,学生的脑力将得到锻炼和提升,这对他们的学习和未来发展都将起到积极的影响。

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爬藤利器名校录取必备!USACO竞赛对参赛者有什么要求?

美国信息学学术活动(USACO)是一项计算机学术活动,类似于国内的全国信息学奥林匹克学术活动(NOIP)。USACO的学术活动分为四个级别:铜级、银级、金级和铂金级。参赛学生从铜级开始,通过晋级来提高自己的学术活动水平。随着级别的提升,学术活动题目的难度也会相应增加。

在USACO学术活动中,学生将面临各种与计算机科学相关的问题。这些问题可能涉及数据结构、算法设计、动态规划、图论等多个领域。参赛学生需要运用自己的知识和技能,分析问题,设计算法,编写代码,并通过优化策略来解决这些问题。

比赛规则

比赛时间为4个小时,中间不能停顿。比赛过程中,看不到测试数据,只有比赛结束后,才能看到测试数据。青铜、白银、黄金、铂金级别的比赛都是3道题,总分1000分。

每道题333.3分。

每道题有10个测试点,通过一个可得33.33分。

铜级是USACO的起点,对于初学者来说是一个很好的入门级别。这个级别的题目相对简单,主要考察基本的编程能力和算法理解。随着学生晋级到更高级别,题目的难度将逐渐增加,需要运用更复杂的算法和更高级的编程技巧来解决。

USACO对参赛者有什么要求?

USACO学术活动对于参赛年龄没有特别限制,只要你是一名高中生,对计算机、编程感兴趣都可以参加这个比赛。

参赛者只需在官网注册即可线上参赛,完全免费。

参加USACO学术活动需要具备一定的编程语言基础。比赛接受以下计算机语言:C++11、Java、C++、Python 3.4.0、Python 2.7.6、C和Pascal。

USACO不仅提供了一个比赛平台,还为参赛选手提供了丰富的学习资源。官方网站上提供了大量的试题、题解、讨论和培训资料,学生可以利用这些资源来提高自己的编程和算法能力。

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USACO竞赛各等级难度如何?晋级到不同级别有何竞争力?

参加USACO学术活动不仅可以在申请美本过程中增加竞争力,还能够掌握重要的计算机算法技能,同时提高学生的脑力和逻辑能力。对于对计算机科学感兴趣的学生来说,参加USACO学术活动无疑是一项值得考虑和积极参与的活动。

USACO的目标是通过学术活动提供一个锻炼计算机编程和算法设计能力的平台,促进学生在计算机科学领域的发展。每年举办的USACO学术活动都吸引了来自世界各地的优秀学生参与其中,他们通过解决一系列挑战性的问题展示自己的才华和技能。

USACO学术活动难度等级

铜级

考试题目相对简单,需要学生至少掌握一种程序语言;

银级

通过铜级考试,需要基本问题解决能力以及算法能力,例如基本数据结构,递归搜索算法等基本算法。

黄金级

通过银级考试,需要有算法基础,掌握高级数据结构,动态规划等高级算法。

白金级

通过黄金级考试,需要很高的编程基础和很强的算法能力,各类高级的数据结构,尤其需要注意算法的时间和空间复杂度。

晋级到不同级别有何竞争力?

白金级别

对于学生来说,如果能够成功晋级到白金级别,这将成为申请名校(如卡内基梅隆大学、佐治亚理工学院和加州大学伯克利分校)时的一大加分项。白金级别的成绩证明了学生在计算机学术活动方面的出色表现和潜力。

黄金级别

对于晋级到黄金级别的学生来说,这是相当不错的成绩。在申请像加州大学伯克利分校、加利福尼亚大学洛杉矶分校和佐治亚理工学院等好学校时,USACO黄金级别的成绩将给予他们额外的加成。这也意味着他们在计算机学术活动中取得了显著的成绩,并且在相关领域显示出了自己的实力。

银级

就算是晋级到银级别,这也是在申请许多大学时的亮点。USACO银级别的成绩证明了学生在计算机学术活动中的积极参与和取得的成就,对于大学招生官来说,这显示了学生的学习能力和成长潜力。

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爬藤利器!USACO竞赛从零基础到入门需要多久?

USACO在国外理工牛校的认可度极高。对于计算机相关专业的学生来说,USACO的晋级和获奖将成为他们申请名校的重要加分项。不论是晋级到白金、黄金还是银级,都代表着学生在计算机学术活动中的成就和潜力。

USACO学术活动提供了一系列的编程问题,旨在培养学生的计算思维和解决问题的能力。在比赛过程中,学生将面临各种难度级别的算法和数据结构问题。通过参加这样的学术活动,学生将有机会提高自己的编程技巧和算法设计能力。

USACO学术活动的难度分为四个级别:铜、银、金和白金。初学者通常从铜级别开始,然后逐渐提升到更高的级别。每个级别都有一系列的题目,学生需要在规定的时间内用编程语言写出正确的解答。这些题目涉及广泛的主题,如搜索、排序、图论等,对学生的编程能力、数学水平、逻辑思维和解决问题的能力都提出了很高的要求。

赛事说明

赛事语言:USACO学术活动支持C++,Java,Pascal,Python,C语言

比赛费用:免费

比赛时间:12月、1月、2月、3月

比赛时长:比赛时长4个小时,中间不能停顿

比赛结果:满分当场晋级,非满分考试结束后公布晋级分数线

比赛分值:比赛设置3道题,总分1000分,每道题333.3分

USACO从零基础到入门需要多久?

对于国内许多小学生而言,他们开始学习编程语言,准备参加信息学学术活动。考虑到这些学生年龄较小,他们需要更多的细节讲解,以及更多的练习和个性化的点评时间。基础编程语言的入门通常需要60小时的课程,每次三小时,约为半年的时间。

然而,对于初中以上的学生来说,他们的理解能力已经相当强了,不需要来回重复许多概念。因此,初中以上的学生学习编程语言,大约需要20小时的课程就足够了。在课后,配合做一些习题,这样就可以掌握算法所需的基本编程语言知识点。

学习编程语言非常重要,因为后续的算法思路和逻辑都需要用代码来表达。家长们可以根据孩子的年龄段,选择最适合他们的学习方式,以便尽快打好编程基础,快速进入算法学习的阶段!

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USACO2023年1月美国计算机奥赛竞赛铜奖组问题三——Moo Operations

Because Bessie is bored of playing with her usual text string where the only characters are 'C,' 'O,' and 'W,' Farmer John gave her Q new strings (1≤Q≤100), where the only characters are 'M' and 'O.' Bessie's favorite word out of the characters 'M' and 'O' is obviously "MOO," so she wants to turn each of the Q strings into "MOO" using the following operations:

Replace either the first or last character with its opposite (so that 'M' becomes 'O' and 'O' becomes 'M').
Delete either the first or last character.
Unfortunately, Bessie is lazy and does not want to perform more operations than absolutely necessary. For each string, please help her determine the minimum number of operations necessary to form "MOO" or output −1 if this is impossible.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line of input contains the value of Q.
The next Q lines of input each consist of a string, each of its characters either 'M' or 'O'. Each string has at least 1 and at most 100 characters.

OUTPUT FORMAT (print output to the terminal / stdout):

Output the answer for each input string on a separate line.

SAMPLE INPUT:

3
MOMMOM
MMO
MOO

SAMPLE OUTPUT:

4
-1
0

A sequence of 4 operations transforming the first string into "MOO" is as follows:

Replace the last character with O (operation 1)
Delete the first character (operation 2)
Delete the first character (operation 2)
Delete the first character (operation 2)
The second string cannot be transformed into "MOO." The third string is already "MOO," so no operations need to be performed.

SCORING:

Inputs 2-4: Every string has length at most 3.
Inputs 5-11: No additional constraints.
Problem credits: Aryansh Shrivastava

USACO2023年1月美国计算机奥赛竞赛铜奖组问题二——Air Cownditioning II

With the hottest recorded summer ever at Farmer John's farm, he needs a way to cool down his cows. Thus, he decides to invest in some air conditioners.

Farmer John's N cows (1≤N≤20) live in a barn that contains a sequence of stalls in a row, numbered 1…100. Cow i occupies a range of these stalls, starting from stall si and ending with stall ti. The ranges of stalls occupied by different cows are all disjoint from each-other. Cows have different cooling requirements. Cow i must be cooled by an amount ci, meaning every stall occupied by cow i must have its temperature reduced by at least ci units.

The barn contains M air conditioners, labeled 1…M (1≤M≤10). The ith air conditioner costs mi units of money to operate (1≤mi≤1000) and cools the range of stalls starting from stall ai and ending with stall bi. If running, the ith air conditioner reduces the temperature of all the stalls in this range by pi (1≤pi≤106). Ranges of stalls covered by air conditioners may potentially overlap.

Running a farm is no easy business, so FJ has a tight budget. Please determine the minimum amount of money he needs to spend to keep all of his cows comfortable. It is guaranteed that if FJ uses all of his conditioners, then all cows will be comfortable.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line of input contains N and M.

The next N lines describe cows. The ith of these lines contains si, ti, and ci.

The next M lines describe air conditioners. The ith of these lines contains ai, bi, pi, and mi.

For every input other than the sample, you can assume that M=10.

OUTPUT FORMAT (print output to the terminal / stdout):

Output a single integer telling the minimum amount of money FJ needs to spend to operate enough air conditioners to satisfy all his cows (with the conditions listed above).

SAMPLE INPUT:

2 4
1 5 2
7 9 3
2 9 2 3
1 6 2 8
1 2 4 2
6 9 1 5

SAMPLE OUTPUT:

10
One possible solution that results in the least amount of money spent is to select those that cool the intervals [2,9], [1,2], and [6,9], for a cost of 3+2+5=10.

Problem credits: Aryansh Shrivastava and Eric Hsu

USACO2023年1月美国计算机奥赛竞赛铜奖组问题一——Leaders

Farmer John has N cows (2≤N≤105). Each cow has a breed that is either Guernsey or Holstein. As is often the case, the cows are standing in a line, numbered 1…N in this order.

Over the course of the day, each cow writes down a list of cows. Specifically, cow i's list contains the range of cows starting with herself (cow i) up to and including cow Ei (i≤Ei≤N).

FJ has recently discovered that each breed of cow has exactly one distinct leader. FJ does not know who the leaders are, but he knows that each leader must have a list that includes all the cows of their breed, or the other breed's leader (or both).

Help FJ count the number of pairs of cows that could be leaders. It is guaranteed that there is at least one possible pair.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains N.
The second line contains a string of length N, with the ith character denoting the breed of the ith cow (G meaning Guernsey and H meaning Holstein). It is guaranteed that there is at least one Guernsey and one Holstein.

The third line contains E1…EN.

OUTPUT FORMAT (print output to the terminal / stdout):

Output the number of possible pairs of leaders.

SAMPLE INPUT:

4
GHHG
2 4 3 4

SAMPLE OUTPUT:

1
The only valid leader pair is (1,2). Cow 1's list contains the other breed's leader (cow 2). Cow 2's list contains all cows of her breed (Holstein).

No other pairs are valid. For example, (2,4) is invalid since cow 4's list does not contain the other breed's leader, and it also does not contain all cows of her breed.

SAMPLE INPUT:

3
GGH
2 3 3

SAMPLE OUTPUT:

2
There are two valid leader pairs, (1,3) and (2,3).

SCORING

Inputs 3-5: N≤100
Inputs 6-10: N≤3000
Inputs 11-17: No additional constraints.
Problem credits: Mythreya Dharani

USACO2023年1月美国计算机奥赛竞赛银奖组问题三——Moo Route

Farmer Nhoj dropped Bessie in the middle of nowhere! At time t=0, Bessie is located at x=0 on an infinite number line. She frantically searches for an exit by moving left or right by 1 unit each second. However, there actually is no exit and after T seconds, Bessie is back at x=0, tired and resigned.

Farmer Nhoj tries to track Bessie but only knows how many times Bessie crosses x=.5,1.5,2.5,…,(N−1).5, given by an array A0,A1,…,AN−1(1≤N≤105, 1≤ Ai≤106, ∑Ai≤106). Bessie never reaches x>N nor x<0.

In particular, Bessie's route can be represented by a string of T=∑N−1i=0Ai Ls and Rs where the ith character represents the direction Bessie moves in during the ith second. The number of direction changes is defined as the number of occurrences of LRs plus the number of occurrences of RLs.

Please help Farmer Nhoj find any route Bessie could have taken that is consistent with A and minimizes the number of direction changes. It is guaranteed that there is at least one valid route.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains N. The second line contains A0,A1,…,AN−1.

OUTPUT FORMAT (print output to the terminal / stdout):

Output a string S of length T=∑N−1i=0Ai where Si is L or R, indicating the direction Bessie travels in during second i. If there are multiple routes minimizing the number of direction changes, output any.

SAMPLE INPUT:

2
2 4

SAMPLE OUTPUT:

RRLRLL
There is only 1 valid route, corresponding to the route 0→1→2→1→2→1→0. Since this is the only possible route, it also has the minimum number of direction changes.

SAMPLE INPUT:

3
2 4 4

SAMPLE OUTPUT:

RRRLLRRLLL

There are 3 possible routes:

RRLRRLRLLL
RRRLRLLRLL
RRRLLRRLLL
The first two routes have 5 direction changes, while the last one has only 3. Thus the last route is the only correct output.

SCORING:

Inputs 3-5: N≤2
Inputs 3-10: T=A0+A1+⋯+AN−1≤5000
Inputs 11-20: No additional constraints.

Problem credits: Brandon Wang and Claire Zhang

USACO2023年1月美国计算机奥赛竞赛银奖组问题二——Following Directions

**Note: The time limit for this problem is 8s, four times the default.**

Farmer John has a big square field split up into an (N+1)×(N+1) (1≤N≤1500) grid of cells. Let cell (i,j) denote the cell in the ith row from the top and jth column from the left. There is one cow living in every cell (i,j) such that 1≤i,j≤N, and each such cell also contains a signpost pointing either to the right or downward. Also, every cell (i,j) such that either i=N+1 or j=N+1, except for (N+1,N+1), contains a vat of cow feed. Each vat contains cow feed of varying price; the vat at (i,j) costs ci,j(1≤ci,j≤500) to feed each cow.

Every day at dinner time, Farmer John rings the dinner bell, and each cow keeps following the directions of the signposts until it reaches a vat, and is then fed from that vat. Then the cows all return to their original positions for the next day.

To maintain his budget, Farmer John wants to know the total cost to feed all the cows each day. However, during every day, before dinner, the cow in in some cell (i,j) flips the direction of its signpost (from right to down or vice versa). The signpost remains in this direction for the following days as well, unless it is flipped back later.

Given the coordinates of the signpost that is flipped on each day, output the cost for every day (with Q days in total, 1≤Q≤1500).

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains N (1≤N≤1500).
The next N+1 lines contain the rows of the grid from top to bottom, containing the initial directions of the signposts and the costs ci,j of each vat. The first N of these lines contain a string of N directions R or D (representing signposts pointing right or down, respectively), followed by the cost ci,N+1. The (N+1)-th line contains N costs cN+1,j.

The next line contains Q (1≤Q≤1500).

Then follow Q additional lines, each consisting of two integers i and j (1≤i,j≤N), which are the coordinates of the cell whose signpost is flipped on the corresponding day.

OUTPUT FORMAT (print output to the terminal / stdout):

Q+1 lines: the original value of the total cost, followed by the value of the total cost after each flip.

SAMPLE INPUT:

2
RR 1
DD 10
100 500
4
1 1
1 1
1 1
2 1

SAMPLE OUTPUT:

602
701
602
701
1501

Before the first flip, the cows at (1,1) and (1,2) cost 1 to feed, the cow at (2,1) costs 100 to feed, and the cow at (2,2) costs 500 to feed, for a total cost of 602. After the first flip, the direction of the signpost at (1,1) changes from R to D, and the cow at (1,1) now costs 100 to feed (while the others remain unchanged), so the total cost is now 701. The second and third flips switch the same sign back and forth. After the fourth flip, the cows at (1,1) and (2,1) now cost 500 to feed, for a total cost of 1501.

SCORING:

Inputs 2-4: 1≤N,Q≤50
Inputs 5-7: 1≤N,Q≤250
Inputs 2-10: The initial direction in each cell, as well as the queries, are uniformly randomly generated.
Inputs 11-15: No additional constraints.
Problem credits: Benjamin Qi