PAT（甲级）2019春季考试(Python实现)

7-1 Sexy Primes

7-1 Sexy Primes
分数 20
作者 陈越
单位 浙江大学Sexy primes are pairs of primes of the form (p, p+6),
so-named since "sex" is the Latin word for "six".
(Quoted from http://mathworld.wolfram.com/SexyPrimes.html)Now given an integer, you are supposed to tell if it is a sexy prime.
Input Specification:Each input file contains one test case. Each case gives a positive integer N (≤108).
Output Specification:For each case, print in a line Yes if N is a sexy prime, then print in the next line the other sexy prime
paired with N (if the answer is not unique, output the smaller number). Or if N is not a sexy prime,
print No instead, then print in the next line the smallest sexy prime which is larger than N.
Sample Input 1:47Sample Output 1:Yes
41Sample Input 2:21Sample Output 2:No
23代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

代码展示:

# Sexy Primes
import math
def isPrime(N):if N <= 1:return Falsefor i in range(2, int(math.sqrt(N)) + 1):if N % i == 0:return Falsereturn Truedef isSexPrime(N):if isPrime(N):if isPrime(N-6) or isPrime(N+6):return Trueelse:return Falseif __name__ == '__main__':N = int(input().strip())if isSexPrime(N):print('Yes')if isPrime(N-6):print(N-6)else:print(N+6)else:print('No')while True:N += 1if isSexPrime(N):print(N)break

7-2 Anniversary

7-2 Anniversary
分数 25
作者 陈越
单位 浙江大学Zhejiang University is about to celebrate her 122th anniversary in 2019.
To prepare for the celebration, the alumni association （校友会） has gathered
the ID's of all her alumni. Now your job is to write a program to count the
number of alumni among all the people who come to the celebration.
Input Specification:Each input file contains one test case. For each case, the first part is about the information of all
the alumni. Given in the first line is a positive integer N (≤105). Then N lines follow, each
contains an ID number of an alumnus. An ID number is a string of 18 digits or the letter X.
It is guaranteed that all the ID's are distinct.The next part gives the information of all the people who come to the celebration.
Again given in the first line is a positive integer M (≤105). Then M lines follow, each
contains an ID number of a guest. It is guaranteed that all the ID's are distinct.
Output Specification:First print in a line the number of alumni among all the people who come to the celebration.
Then in the second line, print the ID of the oldest alumnus -- notice that the 7th - 14th digits of
the ID gives one's birth date. If no alumnus comes, output the ID of the oldest guest instead.
It is guaranteed that such an alumnus or guest is unique.
Sample Input:5
372928196906118710
610481197806202213
440684198612150417
13072819571002001X
150702193604190912
6
530125197901260019
150702193604190912
220221196701020034
610481197806202213
440684198612150417
370205198709275042Sample Output:3
150702193604190912代码长度限制
16 KB
时间限制
800 ms
内存限制
64 MB

代码展示:

# Anniversary
N = int(input().strip())  # 校友数量
alumni = set()
for i in range(N):alumnus = input().strip()alumni.add(alumnus)     M = int(input().strip())  # 访客数量
# 372928196906118710cnt = 0  # 记录参会校友数量
oldestGuest, oldestAlumnus = 'z'*18, 'z'*18
for _ in range(M):guest = input().strip()if guest in alumni:cnt += 1if guest[6:13] < oldestAlumnus[6:13]:oldestAlumnus = guestelse:if guest[6:13] < oldestGuest[6:13]:oldestGuest = guest
print(cnt)
if cnt == 0:print(oldestGuest)
else:print(oldestAlumnus)

7-3 Telefraud Detection

7-3 Telefraud Detection
分数 25
作者 陈越
单位 浙江大学Telefraud（电信诈骗） remains a common and persistent problem in our society.
In some cases, unsuspecting victims lose their entire life savings. To stop this crime,
you are supposed to write a program to detect those suspects from a huge amount of
phone call records.A person must be detected as a suspect if he/she makes more than K short phone calls
to different people everyday, but no more than 20% of these people would call back.
And more, if two suspects are calling each other, we say they might belong to the same gang.
A makes a short phone call to B means that the total duration of the calls from A to B is
no more than 5 minutes.
Input Specification:Each input file contains one test case. For each case, the first line gives 3 positive
integers K (≤500, the threshold（阈值） of the amount of short phone calls), N (≤103,
the number of different phone numbers), and M (≤105, the number of phone call records).
Then M lines of one day's records are given, each in the format:caller receiver durationwhere caller and receiver are numbered from 1 to N, and duration is no more than 1440 minutes
in a day.
Output Specification:Print in each line all the detected suspects in a gang, in ascending order of their numbers.
The gangs are printed in ascending order of their first members. The numbers in a line
must be separated by exactly 1 space, and there must be no extra space at the beginning
or the end of the line.If no one is detected, output None instead.
Sample Input 1:5 15 31
1 4 2
1 5 2
1 5 4
1 7 5
1 8 3
1 9 1
1 6 5
1 15 2
1 15 5
3 2 2
3 5 15
3 13 1
3 12 1
3 14 1
3 10 2
3 11 5
5 2 1
5 3 10
5 1 1
5 7 2
5 6 1
5 13 4
5 15 1
11 10 5
12 14 1
6 1 1
6 9 2
6 10 5
6 11 2
6 12 1
6 13 1Sample Output 1:3 5
6Note: In sample 1, although 1 had 9 records, but there were 7 distinct receivers, among
which 5 and 15 both had conversations lasted more than 5 minutes in total. Hence 1 had
made 5 short phone calls and didn't exceed the threshold 5, and therefore is not a suspect.
Sample Input 2:5 7 8
1 2 1
1 3 1
1 4 1
1 5 1
1 6 1
1 7 1
2 1 1
3 1 1Sample Output 2:None代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

代码展示:(运行超时)

# 7_3 Telefraud Detection
# import collections
K, N, M = map(int, input().strip().split())CallRecordDict = dict()
for _ in range(M):caller, receiver, duration = map(eval, input().strip().split())if caller == receiver:continueCallRecordDict[(caller, receiver)] = CallRecordDict.get((caller, receiver), 0) + duration
#########
shortReceiverDict = dict()
for call, duration in CallRecordDict.items():if duration > 5:  # 长于五分钟的不视为短通话continuecaller, receiver = callif caller not in shortReceiverDict:shortReceiverDict[caller] = [receiver]else:shortReceiverDict[caller].append(receiver)suspects = list()
for caller, receivers in shortReceiverDict.items():shortCallNum = len(receivers)if shortCallNum <= K:  # 短通话数量必须超过Kcontinuecallback_cnt = 0for recv in receivers:if (recv, caller) in CallRecordDict:callback_cnt += 1if callback_cnt <= shortCallNum / 5:suspects.append(caller)suspectsNum = len(suspects)
if suspectsNum == 0:print("None")    quit()HeightsDict, RootsDict = {i : 1 for i in suspects}, {j : j for j in suspects}def getRoot(RootsDict,child):if RootsDict[child] == child:return childelse:root = getRoot(RootsDict, RootsDict[child])RootsDict[child] = rootreturn rootfor i in range(suspectsNum):for j in range(i+1, suspectsNum):gangA, gangB = suspects[i], suspects[j]if ((gangA, gangB) in CallRecordDict) and \((gangB, gangA) in CallRecordDict):  # 两个嫌疑犯相互通话rootA, rootB = getRoot(RootsDict,gangA), getRoot(RootsDict,gangB)if rootA == rootB:continueHeightA, HeightB = HeightsDict[rootA], HeightsDict[rootB]if HeightA > HeightB:RootsDict[rootB] = rootAelif HeightB > HeightA:RootsDict[rootA] = rootBelse:RootsDict[rootB] = rootAHeightsDict[rootA] += 1gangsDict = dict()
for suspect in suspects:root = getRoot(RootsDict, suspect)if root not in gangsDict:gangsDict[root] = [suspect]else:gangsDict[root].append(suspect)result = list()
for k, v in gangsDict.items():v.sort()result.append(v)result.sort()
for item in result:print(*item)

7-4 Structure of a Binary Tree

7-4 Structure of a Binary Tree
分数 30
作者 陈越
单位 浙江大学Suppose that all the keys in a binary tree are distinct positive integers. Given the
postorder and inorder traversal sequences, a binary tree can be uniquely determined.Now given a sequence of statements about the structure of the resulting tree,
you are supposed to tell if they are correct or not. A statment is one of the following:A is the rootA and B are siblingsA is the parent of BA is the left child of BA is the right child of BA and B are on the same levelIt is a full treeNote:Two nodes are on the same level, means that they have the same depth.A full binary tree is a tree in which every node other than the leaves has two children.Input Specification:Each input file contains one test case. For each case, the first line gives a positive
integer N (≤30), the total number of nodes in the binary tree. The second line gives the
postorder sequence and the third line gives the inorder sequence. All the numbers in a
line are no more than 103 and are separated by a space.Then another positive integer M (≤30) is given, followed by M lines of statements. It is
guaranteed that both A and B in the statements are in the tree.
Output Specification:For each statement, print in a line Yes if it is correct, or No if not.
Sample Input:9
16 7 11 32 28 2 23 8 15
16 23 7 32 11 2 28 15 8
7
15 is the root
8 and 2 are siblings
32 is the parent of 11
23 is the left child of 16
28 is the right child of 2
7 and 11 are on the same level
It is a full treeSample Output:Yes
No
Yes
No
Yes
Yes
Yes代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

代码展示:

# 7-4 Structure of a Binary Treeclass TreeNode():def __init__(self, data):self.data = dataself.depth = Noneself.leftChild, self.rightChild = None, Nonedef preOrder(self):print(self.data, end=' ')if self.leftChild != None:self.leftChild.preOrder()if self.rightChild != None:self.rightChild.preOrder()def inOrder(self):if self.leftChild != None:self.leftChild.inOrder()print(self.data, end=' ')    if self.rightChild != None:self.rightChild.inOrder()def postOrder(self):if self.leftChild != None:self.leftChild.postOrder()if self.rightChild != None:self.rightChild.postOrder()print(self.data, end=' ')def addIntoDict(self, treeDict):treeDict[self.data] = selfif self.leftChild != None:self.leftChild.addIntoDict(treeDict)if self.rightChild != None:self.rightChild.addIntoDict(treeDict)def setParent(self, parent):self.parent = parentif self.leftChild != None:self.leftChild.setParent(self)if self.rightChild != None:self.rightChild.setParent(self)def setDepth(self,depth):self.depth = depthif self.rightChild != None:self.rightChild.setDepth(depth+1)if self.leftChild != None:self.leftChild.setDepth(depth+1)def isFullTree(self):if self.rightChild == None == self.leftChild:return Trueelif self.rightChild != None and self.leftChild == None:return Falseelif self.rightChild == None and self.leftChild != None:return Falseelse:  # 左右子树均不为空if self.rightChild.isFullTree() and self.leftChild.isFullTree():return Trueelse:return Falsedef isRoot(self):return self.parent == Nonedef isSiblings(self, other):if self.parent != None:if self.parent == other.parent:return Truereturn Falsedef buildTree(PostSeq, InSeq):if len(PostSeq) == 0:return Nonedata = PostSeq[-1]index = InSeq.index(data)tree = TreeNode(data=data)tree.leftChild = buildTree(PostSeq[:index], InSeq[:index])tree.rightChild = buildTree(PostSeq[index:-1], InSeq[1+index:])return treeN = int(input().strip())
PostSeq = list(map(int, input().strip().split()))
InSeq = list(map(int, input().strip().split()))
tree = buildTree(PostSeq=PostSeq, InSeq=InSeq)
TreeDict = dict()
tree.addIntoDict(TreeDict)
tree.setParent(None)
tree.setDepth(1)M = int(input().strip())for _ in range(M):line = input().strip()if 'root' in line:A = int(line.split()[0])if TreeDict[A].isRoot():print("Yes")else:print("No")elif 'siblings' in line:ls = line.split()A, B = TreeDict[int(ls[0])], TreeDict[int(ls[2])]if A.isSiblings(B):print("Yes")else:print("No")elif 'parent' in line:ls = line.split()A, B = TreeDict[int(ls[0])], TreeDict[int(ls[-1])]if A == B.parent:print("Yes")else:print("No")elif 'left' in line:ls = line.split()A, B = TreeDict[int(ls[0])], TreeDict[int(ls[-1])]if A == B.leftChild:print("Yes")else:print("No")elif 'right' in line:ls = line.split()A, B = TreeDict[int(ls[0])], TreeDict[int(ls[-1])]if A == B.rightChild:print("Yes")else:print("No")elif 'level' in line:ls = line.split()A, B = TreeDict[int(ls[0])], TreeDict[int(ls[2])]if A.depth == B.depth:print("Yes")else:print("No")elif 'full' in line:if tree.isFullTree():print("Yes")else:print("No")

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