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

USACO2023年1月美国计算机奥赛竞赛银奖组问题一——Find and Replace

Bessie is using the latest and greatest innovation in text-editing software, miV! She starts with an input string consisting solely of upper and lowercase English letters and wishes to transform it into some output string. With just one keystroke, miV allows her to replace all occurrences of one English letter c1 in the string with another English letter c2. For example, given the string aAbBa, if Bessie selects c1 as 'a' and c2 as 'B', the given string transforms into BAbBB.

Bessie is a busy cow, so for each of T (1≤T≤10) independent test cases, output the minimum number of keystrokes required to transform her input string into her desired output string.

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

The first line contains T, the number of independent test cases.
The following T pairs of lines contain an input and output string of equal length. All characters are upper or lowercase English letters (either A through Z or a through z). The sum of the lengths of all strings does not exceed 105.

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

For each test case, output the minimum number of keystrokes required to change the input string into the output string, or −1 if it is impossible to do so.

SAMPLE INPUT:

4
abc
abc
BBC
ABC
abc
bbc
ABCD
BACD

SAMPLE OUTPUT:

0
-1
1
3

The first input string is the same as its output string, so no keystrokes are required.

The second input string cannot be changed into its output string because Bessie cannot change one 'B' to 'A' while keeping the other as 'B'.

The third input string can be changed into its output string by changing 'a' to 'b'.

The last input string can be changed into its output string like so: ABCD→EBCD→EACD→BACD.

SCORING:

Inputs 2-6: Every string has a length at most 50.
Inputs 7-9: All strings consist only of lowercase letters 'a' through 'e'
Inputs 10-15: No additional constraints.
Problem credits: Benjamin Qi

牛剑藤录取必备竞赛!USACO竞赛3至12年级学生如何规划?

USACO学术活动是美国计算机奥林匹克学术活动,旨在选拔和培养杰出的计算机科学人才。这项学术活动从初级到高级编程难度逐渐加深,要求参赛者具备扎实的编程基础和深入的计算机理论知识。因此,能够在USACO学术活动中取得好成绩表明申请人在计算机领域有着深厚的才华和知识储备。

在规划备考USACO计算机学术活动时,对于不同年级的学生,需要采取不同的基础规划。下面是一个从3-12年级的规划建议:

3-5年级:

在这个阶段,学生主要需要培养对计算机科学的兴趣和基本的编程思维能力。可以通过参加一些编程俱乐部、夏令营或者在线编程平台上的入门课程来培养孩子的兴趣,如Scratch、Code.org等。了解基本的编程概念和算法原理。

6-8年级:

在这个阶段,学生可以开始系统地学习计算机科学的知识,并开始准备参加USACO青铜级别的学术活动。建议学生选择一门高质量的编程语言作为主力语言,如Python或Java,并学习对应的数据结构和算法。可以通过参加USACO官方提供的练习题和培训班来提高编程能力和学术活动技巧。

9-10年级:

在这个阶段,学生已经掌握了较为扎实的基础知识,在参加USACO学术活动时可以有一定的竞争力。建议学生继续深入学习数据结构和算法,并参加相关的培训班或在线课程来提高编程水平。此外,多参加模拟比赛和解题训练,通过学习和分析他人的解题思路来提高自己的学术活动能力。

11-12年级:

在这个阶段,学生可以进一步提高自己的学术活动水平,并争取达到USACO白金级别。可以选择参加更高级别的培训班或在线课程,学习更复杂的数据结构和算法,进一步提高编程技巧。还可以积极参加USACO的月赛和公开赛,通过与其他优秀选手的切磋来提高自己的学术活动能力。

在整个备考过程中,除了学习编程知识和解题技巧,学生还需要多做练习题,并不断总结经验和找到解题的思路。同时,也要注意与其他学术活动选手交流和分享,多参加相关的讨论社区和比赛活动,扩大自己的视野和认识。

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USACO竞赛报名规则详解!USACO竞赛有哪些注意事项?

留学申请中,USACO学术活动的优秀成绩是一项重要的学术成就。这一成就能够充分展示申请人在计算机科学领域的出色能力和潜力。借助这一突出的学术活动成绩,申请人能够提升留学申请的竞争力,进而增加获得顶级学府录取和奖学金资助的机会。

USACO学术活动报名规则:

参赛者可在官网注册账号,注册 =报名,只需在比赛时间登陆完成答题即可。

注册USACO非常简单,只需要在www.usaco.org网站注册一个免费账户即可。注册时,你不需要选择特定的比赛日期。一旦拥有了USACO账户,你可以在比赛日期随时参加学术活动。如果你已经注册了USACO账户,在考试开放时间内登录你的账户,即可进入比赛。

参赛须知

USACO学术活动每次持续四天,考试时间从周五到周一。学生需要在连续的4小时内参加考试,期间不能暂停。学术活动一共包含3道题目,学生可以反复提交答案,并且提交后会知道有多少个测试用例通过了,但是无法查看每个测试用例的具体情况。

需要注意的是,USACO解题绝对不允许复制他人的代码!在学术活动中,严禁讨论题目内容,也不能抄袭他人的解答。一旦发现这些行为,将会被永久封号处理。因此,在参加USACO学术活动时,务必遵守学术活动的规则和道德准则。

参加USACO学术活动能收获什么?

参加USACO学术活动不仅能够提高学生的编程技能和解决问题的能力,还可以拓宽他们的视野和思维方式。在比赛中,学生们将面临各种各样的编程问题,需要灵活运用自己的知识和技能来解决。这些问题可能涉及到算法设计、数据结构、动态规划、图论等各个方面,需要学生们综合运用多种方法来解决。

通过解决这些问题,学生们可以培养自己的逻辑思维能力、分析问题和解决问题的能力,提高自己的编程素养和创新能力。

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