The 2019 “Orz Panda” Cup Programming Contestacm.xidian.edu.cn/orz-panda/2019/editorial.pdf · T = O(n) Programming Contest Training Base Problem A APA of Orz Pandas Teams passed/tried:

程序设计竞赛实训基地Programming Contest Training Base

The 2019 “Orz Panda” Cup Programming ContestClosing Ceremony

Programming Contest Training Base

Xidian UniversityAug. 21, 2019


Part IEditorial


Problem AAPA of Orz Pandas

▶ Teams passed/tried: 1/4 (10 tries)

▶ Solution:▶ Parse the expression into a tree (using stack explicitly, or recursively)▶ LNR (in-order) traverse the tree and output▶ T = O(n)



▶ Teams passed/tried: 1/4 (10 tries)▶ Solution:

▶ Parse the expression into a tree (using stack explicitly, or recursively)▶ LNR (in-order) traverse the tree and output

▶ T = O(n)




▶ Parse the expression into a tree (using stack explicitly, or recursively)▶ LNR (in-order) traverse the tree and output▶ T = O(n)


Problem BBrute Force of Orz Pandas


▶ Solution:▶ We only want one line so we can determine which branch we should

recursively call▶ Compare k with 2n − 1▶ For n > 100, k < 263 < 2n − 1▶ Swapping another and to, and then swap back...▶ So for n > 100 we can reduce n to n − 2▶ For n = 102, 104, · · · , reduce to n = 100▶ For n = 101, 103, · · · , reduce to n = 99.▶ T = O(1)




▶ We only want one line so we can determine which branch we shouldrecursively call

▶ Compare k with 2n − 1

▶ For n > 100, k < 263 < 2n − 1▶ Swapping another and to, and then swap back...▶ So for n > 100 we can reduce n to n − 2▶ For n = 102, 104, · · · , reduce to n = 100▶ For n = 101, 103, · · · , reduce to n = 99.▶ T = O(1)





▶ Compare k with 2n − 1▶ For n > 100, k < 263 < 2n − 1▶ Swapping another and to, and then swap back...▶ So for n > 100 we can reduce n to n − 2▶ For n = 102, 104, · · · , reduce to n = 100▶ For n = 101, 103, · · · , reduce to n = 99.

▶ T = O(1)





▶ Compare k with 2n − 1▶ For n > 100, k < 263 < 2n − 1▶ Swapping another and to, and then swap back...▶ So for n > 100 we can reduce n to n − 2▶ For n = 102, 104, · · · , reduce to n = 100▶ For n = 101, 103, · · · , reduce to n = 99.▶ T = O(1)


Problem CCai Xukun and Orz Pandas


▶ Solution:▶ It’s easy to see if Cai Xukun has a solution with a velocity v1, he can

always find a solution for all v with |v| > |v1|▶ So we can always use the maximum velocity▶ Solve the parabola equation to get the answer

v2t2 = x2 + (y + 12

gt2)2

▶ Special case: t = 0▶ T = O(1)


Problem CCai Xukun and Orz Pandas


▶ It’s easy to see if Cai Xukun has a solution with a velocity v1, he canalways find a solution for all v with |v| > |v1|

▶ So we can always use the maximum velocity▶ Solve the parabola equation to get the answer

v2t2 = x2 + (y + 12

gt2)2

▶ Special case: t = 0▶ T = O(1)


Problem DDirector of Orz Pandas


▶ Solution:▶ A classical minimum-cut problem▶ Use a maximum flow algorithm to solve it▶ T = O((n + m)2

√k), using highest label push-relabel algorithm


Problem DDirector of Orz Pandas


▶ A classical minimum-cut problem▶ Use a maximum flow algorithm to solve it▶ T = O((n + m)2

√k), using highest label push-relabel algorithm


Problem EExpectation of Orz Pandas


▶ Solution:▶ The answer is (sum of all numbers in[l, r])/(number of stripes in[l, r])▶ Use a segment tree to maintain them▶ Use power series summary to maintain the labels in the tree▶ T = O(m log n)


Problem EExpectation of Orz Pandas


▶ The answer is (sum of all numbers in[l, r])/(number of stripes in[l, r])▶ Use a segment tree to maintain them▶ Use power series summary to maintain the labels in the tree▶ T = O(m log n)


Problem FFullerene of Orz Pandas

▶ Teams passed/tried: 0/0

▶ Solution:▶ Try using Burnside’s lemma:

|X/G| = 1|G|

∑g∈G

c0(g) =1

|G|∑g∈G

∑x∈X

[g · x = x]

▶ What is G?


Problem FFullerene of Orz Pandas

▶ Teams passed/tried: 0/0▶ Solution:▶ Try using Burnside’s lemma:

|X/G| = 1|G|

∑g∈G

c0(g) =1

|G|∑g∈G

∑x∈X

[g · x = x]

▶ What is G?

程序设计竞赛实训基地Programming Contest Training Base Rotations in G

▶ 12× (C5 + C25)▶ 20× C3▶ 15× C2

程序设计竞赛实训基地Programming Contest Training Base Counting c0

▶ Note that 5 is a prime, and there are no points on the C5 axis...

▶ We can see the C5 permutation can be divided into 12 5-cycles.▶ To make the coloring x unchanged with C5 operation, all points

belonging to the same 5-cycle must have same color.▶ We can use 60/5 = 12 colors. So c0(C5) = n12.▶ Similarly, c0(C3) = n20, and c0(C2) = n30.


▶ Note that 5 is a prime, and there are no points on the C5 axis...▶ We can see the C5 permutation can be divided into 12 5-cycles.▶ To make the coloring x unchanged with C5 operation, all points

belonging to the same 5-cycle must have same color.▶ We can use 60/5 = 12 colors. So c0(C5) = n12.

▶ Similarly, c0(C3) = n20, and c0(C2) = n30.


▶ Note that 5 is a prime, and there are no points on the C5 axis...▶ We can see the C5 permutation can be divided into 12 5-cycles.▶ To make the coloring x unchanged with C5 operation, all points

belonging to the same 5-cycle must have same color.▶ We can use 60/5 = 12 colors. So c0(C5) = n12.▶ Similarly, c0(C3) = n20, and c0(C2) = n30.

程序设计竞赛实训基地Programming Contest Training Base At last

|X/G| = n60 + 12× 2× n12 + 20× n20 + 15× n30

1 + 12× 2 + 20 + 15

▶ T = O(1) (or for lazy people T = O(log n), with a naive Pythonprogram)


Problem GGame of Orz Pandas

▶ Teams passed/tried: 1/1 (6 tries)▶ Use SG theorem at first, we just need to enumerate the ways to

make the xor of all selected numbers is exactly x.▶ Then use Walsh transform. Then we only need to multiply the

numbers, then do a reverse Walsh transform to get the answer.▶ Prefix product - just like prefix sum:∏r

k=l Yk = prod(r) · prod(l − 1)−1.

程序设计竞赛实训基地Programming Contest Training Base A special case

▶ Special case: Yk ≡ 0.

▶ We can check if there is an Yk ≡ 0 where k ∈ [l, r] using a prefixsum, and return 0 for these cases.

▶ Then we ignore the numbers ≡ 0 in prefix product, so they won’taffect other queries.

▶ We we do the reverse Walsh transform, since we only need y(x)instead of the entire y, we should use the naive algorithm (in O(P)),instead of O(P log P) FWT. (Q = max{pi, x}.)

▶ T = O(nP log P + QP) (with FWT preprocessing) orT = O(P2 + mP + QP) (with naive preprocessing)

程序设计竞赛实训基地Programming Contest Training Base A special case

▶ Special case: Yk ≡ 0.▶ We can check if there is an Yk ≡ 0 where k ∈ [l, r] using a prefix

sum, and return 0 for these cases.▶ Then we ignore the numbers ≡ 0 in prefix product, so they won’t

affect other queries.▶ We we do the reverse Walsh transform, since we only need y(x)

instead of the entire y, we should use the naive algorithm (in O(P)),instead of O(P log P) FWT. (Q = max{pi, x}.)

▶ T = O(nP log P + QP) (with FWT preprocessing) orT = O(P2 + mP + QP) (with naive preprocessing)


Problem HHorton and Orz Pandas

▶ Teams passed/tried: 4/10 (77 tries)▶ At first notice this is a classical rational programming problem.▶ If the optimal solution is k0 = max{B(x)/A(x)}, then we can define

F(x, k) = B(x)− k · A(x)

▶ With a given k, we can solve

∆(k) = maxx

F(x, k)

▶ For optimal solution we have ∆(k0) = maxx F(x, k0) = 0▶ If k1 > k2,

∆(k1) = maxx F(x, k1) = F(xopt1 , k1)

< F(xopt1 , k2) ≤ maxx F(x, k2) = ∆(k2)

程序设计竞赛实训基地Programming Contest Training Base Binary search

▶ So ∆(k) is monotonic. Binary search to find k0.▶ How to evaluate ∆(k) = maxx F(x, k)?▶ Let the weight of the edges to be bi − k · ai, then use Prim’s

algorithm!▶ Special case: for bi − k · ai > 0, we should always add this edge.


Problem IIPv6 Address of Orz Pandas

▶ Teams tried/passed: 3/5 (19 tries)▶ Stupid problem (that means the problem setter is too stupid).▶ Use Python to convert this large integer to hexadecmial ("%x" %

addr)▶ Then find the consecutive zeros with the number of sections as large

as possible. Since the ASCII of : is larger than 0, if there aremultiple solutions choose the zeros near the tail of the string.


Problem JJobless Orz Pandas

▶ Teams passed/tried: 1/1 (1 try)▶ The signed volumn is multiplied by a factor of |A| if the coordinates

are transformed by matrix A.▶ So the answer is (|A|−1

∏bi)2.


Part IIAwards

程序设计竞赛实训基地Programming Contest Training Base Fastest to Solve

▶ Geometric Rhythm: Fastest to solve problem B, C, D, E, G, and J▶ Two old doges: Fastest to solve problem H▶ @21:30: Fastest to solve problem A▶ International group of qkoqhh: Fastest to solve problem I

程序设计竞赛实训基地Programming Contest Training Base Bronze Medals

▶ KFC basketball team: 1/135▶ No team name: 1/124▶ Sakura tree: 1/105▶ Output illegal: 1/76

程序设计竞赛实训基地Programming Contest Training Base Silver Medals

▶ Dawn: 2/361▶ WaAutomation: 2/263▶ International group of qkoqhh: 3/698▶ Oooooooor2: 3/607


Gold Medal3rd Place

@21:30: 4/1266▶ Team members: Xie Mingzhe, Hu Guangpeng, and Li Xuanzhe


Gold Medal2nd Place

Two old doges: 4/1767▶ Team members: Dai Yuquan, Li Hongzheng, and Chai Dongchen


Gold MedalThe Champion

Geometric Rhythm: 8/1734▶ Team members: Wang Xiaoqing, Wang Junyi, and Huang Haitong


GL and HF!

EditorialAPA of Orz PandasBrute Force of Orz PandasCai Xukun And Orz PandasDirector of Orz PandasExpectation of Orz PandasFullerene of Orz PandasGame of Orz PandasHorton and Orz PandasIPv6 Address of Orz PandasJobless Orz Pandas

Awards

Documents

The 2019 “Orz Panda” Cup Programming Contestacm.xidian.edu.cn/orz-panda/2019/editorial.pdf · T = O(n) Programming Contest Training Base Problem A APA of Orz Pandas Teams passed/tried: