2025年USACO公开赛银奖组问题二—Compatible Pairs

Deep in the countryside, Farmer John’s cows aren’t just ordinary farm animals—they are part of a clandestine bovine intelligence network. Each cow carries an ID number, carefully assigned by the elite cow cryptographers. However, due to Farmer John's rather haphazard tagging system, some cows ended up with the same ID.

Farmer John noted that there are N(1≤N≤2⋅105) unique ID numbers, and for each unique ID di (0≤di≤109), there are ni(1≤ni≤109) cows who shared it.

The cows can only communicate in pairs, and their secret encryption method has one strict rule: two cows can only exchange information if they are not the same cow and the sum of their ID numbers equals either A or B(0≤A≤B≤2⋅109). A cow can only engage in one conversation at a time (i.e., no cow can be part of more than one pair).

Farmer John would like to maximize the number of disjoint communication pairs to ensure the best information flow. Can you determine the largest number of conversations that can happen at once?

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

The first line contains N, A, B.

The next N lines each contain ni and di. No two di are the same.

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

The maximum number of disjoint communicating pairs that can be formed at the same time.

Note that the large size of integers involved in this problem may require the use of 64-bit integer data types (e.g., a "long long" in C/C++).

SAMPLE INPUT:

4 4 5
17 2
100 0
10 1
200 4

SAMPLE OUTPUT:

118

A cow with an ID of 0 can communicate with another cow with an ID of 4 because the sum of their IDs is 4. Since there are a total of 100 cows of ID 0
and 200 cows of ID 4, there can be up to 100 communicating pairs with this combination of IDs.

A cow with an ID of 4 can also communicate with another cow with an ID of 1
(sum to 5). There are 10 cows of ID 1 and 100 remaining unpaired cows of ID 4
, allowing for another 10 pairs.

Finally, a cow with an ID of 2 can communicate with another cow of the same ID. Since there are a total of 17 cows of ID 2, there can be up to 8 more pairs.

In total, there are 100+10+8=118 communicating pairs. It can be shown that this is the maximum possible number of pairs.

SAMPLE INPUT:

4 4 5
100 0
10 1
100 3
20 4

SAMPLE OUTPUT:

30

Pairing IDs 0 and 4 makes 20 pairs, while pairing IDs 1 and 3 makes 10 pairs. It can be shown that this is the optimal pairing, resulting in a total of 30
pairs.

SCORING:

Inputs 3-4: A=B
Inputs 5-7: N≤1000
Inputs 8-12: No additional constraints

Problem credits: Benjamin Qi

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2025年USACO公开赛银奖组问题一—Sequence Construction

Lately, the cows on Farmer John's farm have been infatuated with watching the show Apothecowry Dairies. The show revolves around a clever bovine sleuth CowCow solving problems of various kinds. Bessie found a new problem from the show, but the solution won't be revealed until the next episode in a week! Please solve the problem for her.

You are given integers M and K (1≤M≤109,1≤K≤31). Please choose a positive integer N and construct a sequence a of N non-negative integers such that the following conditions are satisfied:

1≤N≤100
a1+a2+⋯+aN=M
popcount(a1)⊕ popcount(a2)⊕⋯⊕ popcount(aN)=K
If no such sequence exists, print −1.

† popcount(x) is the number of bits equal to 1 in the binary representation of the integer x. For instance, the popcount of 11 is 3 and the popcount of 16 is 1.

†⊕is the bitwise xor operator.

The input will consist of T (1≤T≤5⋅103) independent test cases.

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

The first line contains T.

The first and only line of each test case has M and K.

It is guaranteed that all test cases are unique.

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

Output the solutions for T test cases as follows:

If no answer exists, the only line for that test case should be −1.

Otherwise, the first line for that test case should be a single integer N, the length of the sequence -- (1≤N≤100).

The second line for that test case should contain N space-separated integers that satisfy the conditions -- (0≤aiM).

SAMPLE INPUT:

3
2 1
33 5
10 5

SAMPLE OUTPUT:

2
2 0
3
3 23 7
-1

In the first test case, the elements in the array a=[2,0] sum to 2. The xor sum of popcounts is 1⊕0=1. Thus, all the conditions are satisfied.

In the second test case, the elements in the array a=[3,23,7] sum to 33. The xor sum of the popcounts is 2⊕4⊕3=5. Thus, all conditions are satisfied.

Other valid arrays are a=[4,2,15,5,7] and a=[1,4,0,27,1].

It can be shown that no valid arrays exist for the third test case.

SCORING:

Input 2: M≤8,K≤8
Inputs 3-5: M>2K
Inputs 6-18: No additional constraints.

Problem credits: Aakash Gokhale

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2025年USACO公开赛金奖组问题三—OohMoo Milk

Farmer John is trying to make his world's famous OohMoo Milk to sell for a profit. He has N (1≤N≤105) bottles that he is trying to fill. Each bottle initially contains some amount of milk mi (0≤mi≤109). Every day, he takes A (1≤AN)
bottles and fills each bottle with one unit of milk.

Unfortunately, Farmer Nhoj, Farmer John's competitor in the business of OohMoo Milk, knows about Farmer John's production processes and has a plan to curtail his business. Every day, after Farmer John fills his A bottles, Farmer Nhoj will sneakily steal one unit of milk from each of B (0≤B<A) different nonempty bottles. To remain sneaky, Farmer Nhoj chooses B so that it is strictly less than A, so that it is less likely for Farmer John to discover him.

After D(1≤D≤109) days, Farmer John will sell his OohMoo Milk. If a bottle has M
units of milk, it will sell for M2 moonies.

Let P be the unique profit such that FJ can guarantee that he makes at least P profit regardless of how FN behaves, and FN can guarantee that FJ makes at most P profit regardless of how FJ behaves. Output the value of P modulo 109+7.

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

The first line of the input contains N and D, where N is the number of bottles and D is the number of days that take place.

The second line of the input contains A and B representing the number of units of milk that Farmer John fills and Farmer Nhoj steals respectively.

The third line of the input contains N space-separated integers mi representing the initial amount of milk in each bottle.

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

Output the value of P modulo 109+7.

SAMPLE INPUT:

5 4
4 2
4 10 8 10 10

SAMPLE OUTPUT:

546

On the first day, Farmer John could add milk to the second, third, fourth, and fifth bottles. Then, Farmer Nhoj could remove milk from the second and fourth bottles.

Thus, the new amount of milk in each bottle is

[4,10,8,10,10]→[4,11,9,11,11]→[4,10,9,10,11].

After four days, the amount of milk in each bottle could be

[4,10,8,10,10]→[4,10,9,10,11]→[4,10,10,11,11]→[4,11,11,11,11]→[4,11,11,12,12].

The total amount of moonies Farmer John would make in this situation is 42+112+112+122+122=546. It can be shown that this is the value of P.

SAMPLE INPUT:

10 5
5 1
1 2 3 4 5 6 7 8 9 10

SAMPLE OUTPUT:

777

SAMPLE INPUT:

5 1000000000
3 1
0 1 2 3 4

SAMPLE OUTPUT:

10

Make sure you output P modulo 109+7.

SCORING:

Inputs 4-6: N,D≤1000.
Inputs 7-10: D≤106.
Inputs 11-20: No additional constraints.

Problem credits: Suhas Nagar

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2025年USACO公开赛金奖组问题二—Election Queries

**Note: The time limit for this problem is 3s, 1.5x the default.**

Farmer John has N(2≤N≤2⋅105) cows numbered from 1 to N. An election is being held in FJ's farm to determine the two new head cows in his farm. Initially, it is known that cow i will vote for cow ai  (1≤ai ≤N).

To determine the two head cows, FJ will hold his election in the following process:

Choose an arbitrary subset S of his cows that contains at least one cow but not all cows. FJ is able to choose cow x as the first head cow if its votes appear most frequently among all votes cast by cows in S.
FJ is able to choose cow y as the second head cow if its votes appear most frequently among votes cast by cows not in S.
For a fixed subset S, FJ denotes the diversity between his head cows as |xy|. As FJ does not like having leaders with similar numbers, he wants to choose S
such that the diversity is maximized. Note that if FJ is not able to choose two distinct head cows, then the diversity is 0.

However, some cows keep changing their minds, and FJ may have to rerun the election many times! Therefore, he asks you Q (1≤Q≤105) queries. In each query, a cow changes their vote. After each query, he asks you for the maximum possible diversity among his new head cows.

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

The first line contains N and Q.

The following line contains a1,a2,…,aN.

The following Q lines contain two integers i and x, representing the update ai=x
(1≤i,xN).

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

Output Q lines, the i'th of which is the maximum possible diversity after the first i
queries.

SAMPLE INPUT:

5 3
1 2 3 4 5
3 4
1 2
5 2

SAMPLE OUTPUT:

4
3
2

After the first query, a=[1,2,4,4,5]. At the first step of the election, FJ can make S={1,3}. Here, cow 1 receives one vote and cow 4 receives one vote. Therefore, FJ can choose either cow 1 or cow 4 as its first head cow.

For all cows not in the election, cow 2 receives one vote, cow 4 receives one vote, and cow 5 also receives one vote. Therefore, FJ can choose any one of cows 2, 4,or 5 to be its second head cow.

To obtain the maximum diversity, FJ can choose cow 1 as the first head cow and cow 5 as the second head cow. Therefore, the diversity is |1−5|=4.

After the second query, a=[2,2,4,4,5] and FJ can make S={4,5}. Then, he can choose 5 as the first head cow and cow 2 as the second head cow. The maximum possible diversity is |5−2|=3.

SAMPLE INPUT:

8 5
8 1 4 2 5 4 2 3
7 4
8 4
4 1
5 8
8 4

SAMPLE OUTPUT:

4
4
4
7
7

SCORING:

Inputs 3-4: N,Q≤100
Inputs 5-7: N,Q≤3000
Inputs 8-15: No additional constraints.
Problem credits: Chongtian Ma and Haokai Ma

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2025年USACO公开赛金奖组问题一—Moo Decomposition

You have a long string S of Ms and Os and an integer K≥1. Count the number of ways of ways to decompose S into subsequences such that each subsequence is MOOOO....O with exactly K Os, modulo 109+7.

Since the string is very long, you are not given it explicitly. Instead, you are given an integer L (1≤L≤1018), and a string T of length N (1≤N≤106). The string S is the concatenation of L copies of the string T.

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

The first line contains K, N, and L.
The second line contains the string T of length N. Every character is either an M or an O.

It is guaranteed that the number of decompositions of S is nonzero.

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

Output the number of decompositions of string S, modulo 109+7.

SAMPLE INPUT:

2 6 1
MOOMOO

SAMPLE OUTPUT:

1

The only way to decompose S into MOOs is to let the first three characters form a MOO and the last three characters form another MOO.

SAMPLE INPUT:

2 6 1
MMOOOO

SAMPLE OUTPUT:

6

There are six distinct ways to decompose the string into subsequences (uppercase letters form one MOO, lowercase letters form another):

MmOOoo
MmOoOo
MmOooO
MmoOOo
MmoOoO
MmooOO

SAMPLE INPUT:

1 4 2
MMOO

SAMPLE OUTPUT:

4

SAMPLE INPUT:

1 4 100
MMOO

SAMPLE OUTPUT:

976371285

Make sure to take the answer modulo 109+7.

SCORING:

Inputs 5-7: K=1, L=1
Inputs 8-10: K=2, N≤1000, L=1
Inputs 11-13: K=1
Inputs 14-19: L=1
Inputs 20-25: No additional constraints.

Problem credits: Dhruv Rohatgi

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2025年USACO公开赛白金奖组问题三—Package Pickup

**Note: The time limit for this problem is 4s, 2x the default.**

Farmer John has distributed cows and packages in a weird pattern across the number line using the following process:

Farmer John chooses a number M (1≤M≤1018).
Farmer John chooses N (1≤N≤2⋅104) intervals [ Li,Ri] to distribute cows in (1≤LiRi≤1018). He then places cows at locations Li,Li+M,Li+2M,…,Ri. It is guaranteed that RiLi is a multiple of M.
Farmer John chooses P (1≤P≤2⋅104) intervals [Ai,Bi] to distribute packages in (1≤AiBi≤1018). He then places packages at locations Ai ,Ai +M,Ai +2M,…,Bi
. It is guaranteed that BiAi is a multiple of M.

Once the cows and packages are distributed, Farmer John wants to see how long it takes the cows to pick up the packages. Every second, Farmer John can issue a command to a single cow to move one unit left or right of their current position with his handy walkie talkie. If a cow travels to the position where a package is located, they are able to pick it up. Farmer John wants to know the minimum time in seconds that it would take the cows to pick up every package.

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

The first line contains M, N, and P.

The next N lines each contain two integers Li and Ri.

The next P lines each contain two integers Ai and Bi.

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

Output a single integer, representing the minimum amount of time it can take the cows to pick up all the packages, given that every second, he can issue a single left/right command to a single cow.

SAMPLE INPUT:

100 3 7
10 10
20 20
30 30
7 7
11 11
13 13
17 17
24 24
26 26
33 33

SAMPLE OUTPUT:

22

In the above test case, suppose the cows and packages are numbered from left to right. Farmer John can follow this procedure to pick up the packages in 22 seconds:

Issue 3 lefts to cow 1 so that it picks up package 1
Issue 3 rights to cow 3 so that it picks up package 7
Issue 4 rights to cow 2 so that it picks up package 5
Issue 10 rights to cow 1 so that it picks up packages 2, 3, and 4
Issue 2 rights to cow 2 so that it picks up package 6

SAMPLE INPUT:

2 1 1
1 5
2 6

SAMPLE OUTPUT:

3

There are three cows and three packages. Farmer John can issue one right to each cow.

SCORING:

Input 3-4: It is guaranteed that the total number of cows and packages does not exceed 2⋅105 
Inputs 5-10: It is guaranteed that N,P≤500.
Inputs 11-13: It is guaranteed that no intervals of packages or cows intersect.
Inputs 14-20: No additional constraints.

Problem credits: Suhas Nagar and Benjamin Qi

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2025年USACO公开赛白金奖组问题二—Lazy Sort

Farmer John has N cows (2≤N≤5⋅106) and is attempting to get them to sort a non-negative integer array A of length N by relying on their laziness. He has a lot of heavy boxes so he lines the cows up one behind another, where cow i+1 is behind cow i, and gives ai boxes to cow i (0≤ai).

Cows are inherently lazy so they always look to pass their work off to someone else. From cow 1 to N−1 in order, each cow looks to the cow behind them. If cow i
has strictly more boxes than cow i+1, cow i thinks this is "unfair" and gives one of its boxes to cow i+1. This process repeats until every cow is satisfied.

Farmer John will then note the number of boxes bi that each cow i is holding and create an array B out of these values. If B=sorted(A), then Farmer John will be happy. Unfortunately, Farmer John forgot all but Q values (2≤Q≤min(N,100)) in A. Luckily, those values include the number of boxes he was going to give to the first and last cow. Each value that FJ remembers is given in the form ci vi representing that a ci=vi(1≤ci≤N, 1≤vi ≤109). Determine the number of different ways the missing values can be filled in so that he will be happy mod 109+7.

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

The first line contains two space-separated integers N and Q representing the number of cows and queries respectively.

The next Q lines contain two space separated integers ci vi representing that cow ci initially holds vboxes. It is guaranteed that c1=1, cQ=N, and ci<ci+1
(the order of the cows is strictly increasing).

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

Print the number of different ways modulo 109+7 that values ai can be assigned such that Farmer John will be happy after the cows perform the lazy sort. It is guaranteed that there will be at least one valid assignment.

SAMPLE INPUT:

3 2
1 3
3 2

SAMPLE OUTPUT:

2

In this example, FJ remembers the values at the ends of the array. The arrays [3,2,2] and [3,3,2] are the valid arrays that will make FJ happy at the end of the lazy sorting.

SAMPLE INPUT:

6 3
1 1
3 3
6 5

SAMPLE OUTPUT:

89

SCORING:

Inputs 3-4: N,vi≤100
Inputs 5-6: N≤100 and vi≤106
Inputs 7-9: N≤2⋅105 and vi≤106
Inputs 10-12: N≤2⋅105
Inputs 13-15: No additional constraints.

Problem credits: Suhas Nagar

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2025 年美国USACO公开赛——最终结果

2025 年美国公开赛以算法编程问题为特色,涵盖广泛的技术和难度级别。

在 5782 天的比赛中,共有 4 名不同的用户登录了比赛。共有 4024 名参与者提交了至少一个解决方案,来自 100+ 个不同的国家/地区。2349 名参与者来自美国,其中来自中国、加拿大、韩国、罗马尼亚、印度和新加坡的参与人数也很高。

总共有 10638 篇评分的提交,按语言细分如下:

6698 C++17

1701 Python-3.6.9

1315 Java

866 C++11

41 C

17 Python-2.7.17

以下是白金、黄金、白银和铜奖比赛的详细结果。 您还可以找到每个问题的解决方案和测试数据。

USACO 2025 年 美国公开赛,白金奖

白金组共有 255 名参与者,其中 199 名是大学预科生。得分最高的成员的成绩在这里。祝贺所有最优秀的参赛者取得优异的成绩!

1 Forklift Certified
查看问题 | 测试数据 | 解决方案
2 Lazy Sort
查看问题 | 测试数据 | 解决方案
3 Package Pickup
查看问题 | 测试数据 | 解决方案

USACO 2025 年美国公开赛,金牌

黄金组共有 856 名参与者,其中 627 名是大学预科生。所有在本次比赛中获得 850 分或更高分的参赛者将自动晋升到白金组。获得晋升的美国大学预科生名单在这里

1 Moo Decomposition
查看问题 | 测试数据 | 解决方案
2 Election Queries
查看问题 | 测试数据 | 解决方案
3 OohMoo Milk
查看问题 | 测试数据 | 解决方案

USACO 2025 年美国公开赛,银牌

白银组共有 2000 名参与者,其中 1545 名是大学预科生。所有在本次比赛中获得 750 分或更高分的参赛者将自动晋级黄金组。

1 Sequence Construction
查看问题 | 测试数据 | 解决方案
2 Compatible Pairs
查看问题 | 测试数据 | 解决方案
3 Ski Slope
查看问题 | 测试数据 | 解决方案

USACO 2025 年美国公开赛,铜牌

铜牌组共有 2461 名参与者,其中 1883 名是大学预科生。所有在本次比赛中获得 700 分或更高分的参赛者将自动晋升到银牌组。

1 Hoof Paper Scissors Minus One
查看问题 | 测试数据 | 解决方案
2 More Cow Photos
查看问题 | 测试数据 | 解决方案
3 It's Mooin' Time III

查看问题 | 测试数据 | 解决方案

结语

2024-2025 赛季现已结束。我很高兴看到,尽管问题重重,但本赛季的所有比赛都进行得很顺利,并带来了大量的晋升。祝贺所有参与其中的人,他们的编码和解决问题的能力在整个赛季都有所提高。

对于那些尚未晋升的人,请记住,你练习得越多,你的算法编码技能就会越好——请坚持下去!USACO竞赛旨在挑战最优秀的学生,要想在他们职中脱颖而出,可能需要付出大量的努力。为了帮助您修复代码中的任何错误,您现在可以使用“分析模式”重新提交解决方案并从评判服务器获得反馈。

大量的人为USACO竞赛的质量和成功做出了贡献。本次竞赛的协助人员包括Suhas Nagar, Nathan Wang, Nick Wu, William Lin, Chongtian Ma, Alex Liang, Aakash Gokhale, Weiming Zhou, Ho Tin Fan, Michelle Wei, Jichao Qian, Brandon Wang, Dhruv Rohatgi, Thomas Liu, Haokai Ma, Larry Xing, Eric Yang, Austin Geng, Andi Qu, Benjamin Chen, Anand John, and Benjamin Qi。 也感谢我们的翻译帮助我们扩大比赛的范围。最后,我们非常感谢USACO赞助商的慷慨支持:Citadel, 它已成为我们营地和IOI/EGOI团队的独家赞助商,以及Jump Trading,它帮助赞助我们的在线比赛。

祝我们在 2025 年 IOI 和 EGOI 上好运,我们期待着 在 2025-2026 赛季再次见到大家!

祝您编码愉快!

2025年USACO公开赛白金奖组问题一—Forklift Certified

Farmer John is training to become forklift certified! As part of his training, he needs to clear N(1≤N≤105) boxes, conveniently labeled 1 through N, from an old warehouse.

The boxes can be modeled as axis-aligned rectangles in a 2-dimensional plane, where the +x-direction is east and the +y-direction is north. Box i has its southwest corner at (xi1,yi1) and its northeast corner at (xi2,yi2). All coordinates are integers in the range [1,2N], and no two corners from two different rectangles share the same x or y coordinate. All boxes have a non-zero area, and no two boxes intersect.

Farmer John plans to remove the boxes one at a time out of the southwest entrance of the warehouse. However, he can only remove a box if no part of any other box lies both south and west of the box's northeast corner due to physical limitations of the forklift.

An example with N=4 is shown below. To remove box 4, there cannot be any other boxes in the shaded region. Boxes 2 and 3 prevent box 4 from being removed, but box 1 does not.

Help Farmer John decide how to remove all the boxes! Your code should operate in two separate modes, defined by an integer flag M:

Mode 1 (M=1): Generate a permutation of 1,…,N specifying a valid box removal order. If there are multiple valid orders, find any. It can be proven that such an order always exists.
Mode 2 (M=2): For each k=1,…,N, output 1 if Farmer John can remove box k if boxes 1,…,k−1 have already been removed, and 0 otherwise.

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

Each input consists of T (1≤T≤10) independent test cases. It is guaranteed that the sum of all N within each input does not exceed 5⋅105.

The first line of input contains T and M. (Note that M is the same for each test case.) Each test case is then formatted as follows:

The first line contains a single integer N.

Each of the next N lines contains four space-separated integers xi1,yi1, xi2,yi2 the locations of the southwest and northeast corners of box i.

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

For each test case:

If M=1, output a single line with N space-separated integers, where the j-th integer is the label of the j-th box to remove.

If M=2, output a single line with a binary string of N characters specifying the answer for each k=1,…,N.

SAMPLE INPUT:

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

SAMPLE OUTPUT:

1 3 2 4
2 3 1

The first test case corresponds to the N=4 exammple above. Box 1 is not blocked by anything, box 3 is blocked by box 1, box 2 is blocked by box 3, and box 4 is blocked by boxes 2 and 3.

SAMPLE INPUT:

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

SAMPLE OUTPUT:

1011
011

For the first test case, box 2 is blocked by box 3, so Farmer John cannot remove it before removing box 3.

SCORING:

Inputs 3-5: M=1, N≤1000.
Input 6: M=2, N≤1000.
Inputs 7-13: M=1, no additional constraints.
Inputs 14-16: M=2, no additional constraints.

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STEM申请王炸!USACO不同等级在大学申请中有什么优势?

作为一项全球知名的计算机竞赛,USACO的影响力无疑是显著的。许多美国名校,包括麻省理工学院(MIT)、斯坦福大学、哈佛大学等,都将USACO赛事视为了解申请者能力的重要标准。相较于传统的数学竞赛,参加USACO可以更为精准地证明申请者在计算机科学的实践能力,这为有志于在相关领域深造的学生提供了一个更具说服力的背景。

USACO(美国计算机奥林匹克竞赛)根据参赛者的表现分为多个等级,每个等级在大学申请中都有不同的作用和优势。

一、IOI金牌(国际信息学奥林匹克竞赛金牌)

作用:

顶尖名校的“敲门砖”: 获得IOI金牌意味着你在计算机科学领域达到了 世界顶尖水平,这几乎是 保证被MIT(麻省理工学院)、Stanford(斯坦福大学)、Harvard(哈佛大学) 等顶尖名校录取的 “金钥匙”。

全球认可: IOI是全球最具影响力的信息学竞赛,其金牌得主在学术界和工业界都备受瞩目。

申请优势:

几乎确保录取: 获得IOI金牌的学生在申请上述顶尖名校时,几乎可以 确保录取。

奖学金机会: 许多顶尖大学会为IOI金牌得主提供 全额奖学金 和 其他优厚待遇。

二、USACO国家集训队

作用:

顶尖大学申请的“助推器”: 进入USACO国家集训队在申请 MIT、Stanford、Princeton(普林斯顿大学) 等顶尖大学时具有 非常明显和有效的助推作用。

学术能力证明: 表明你在计算机科学和编程方面拥有 卓越的能力 和 潜力。

申请优势:

强有力的学术证明: 国家集训队成员的身份是 强有力的学术证明,可以 增强 申请材料的 竞争力。

面试机会: 一些顶尖大学可能会为USACO国家集训队成员提供 面试机会 或 优先考虑。

三、USACO Platinum Division(铂金级)

作用:

名校申请的“加分项”: 进入USACO铂金级在申请 CMU(卡内基梅隆大学)、Georgia Tech(佐治亚理工学院)、UC Berkeley(加州大学伯克利分校) 等名校时是 很大的加分项。

编程能力体现: 表明你具备 顶尖的编程能力 和 算法设计能力。

申请优势:

学术竞争力: 铂金级成绩可以 显著提升 申请者在 计算机科学 和 工程 等相关专业的 学术竞争力。

奖学金机会: 一些大学可能会为铂金级选手提供 奖学金 或 其他奖励。

四、USACO Gold Division(黄金级)

作用:

好学校申请的“亮点”: 进入USACO黄金级在申请 UC Berkeley、UCLA(加州大学洛杉矶分校)、GIT(佐治亚理工学院) 等好学校时是一个 亮点。

编程能力证明: 表明你具备 优秀的编程能力 和 问题解决能力。

申请优势:

竞争力提升: 黄金级成绩可以 提升 申请者在 顶尖大学 和 热门专业 中的 竞争力。

项目参与机会: 一些大学可能会邀请黄金级选手参与 研究项目 或 实习机会。

五、USACO Silver Division(银级)

作用:

申请亮点: 进入USACO银级在申请 许多大学 时是一个 亮点,可以展示你的 编程兴趣 和 学习能力。

基础能力证明: 表明你具备 良好的编程基础 和 逻辑思维能力。

申请优势:

差异化竞争力: 银级成绩可以帮助申请者在 众多申请者 中 脱颖而出。

课外活动证明: 银级成绩可以作为 课外活动 的一个 有力证明,展示你的 全面发展。

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