Question

我曾尝试过Google的kickstart 2020挑战赛。 C轮问题1让我有些难过。我尝试了许多不同的方式来完成挑战。这个问题看起来很简单，但我无法完成。问题是我不明白自己在做什么错。请为我指出正确的方向，或为我的代码指出问题。

问题

Google Kickstart 2020-C轮|问题1 https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff43/00000000003380d2

Avery has an array of N positive integers. The i-th integer of the array is Ai.

A contiguous subarray is an m-countdown if it is of length m and contains the integers m, m-1, m-2, ..., 2, 1 in that order. For example, [3, 2, 1] is a 3-countdown.

Can you help Avery count the number of K-countdowns in her array?

Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the integers N and K. The second line contains N integers. The i-th integer is Ai.

Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of K-countdowns in her array.

伪代码

    get the number of cases
    Loop in range(number of cases):
        get the N (number of elements), K(initial countdown value)
        get the array of values
        generate an array of the countdown sequence [K ... 1] - signature
        counter = 0

        Loop elem in range(Number of elements):
            if elem == K:
                if there is space to slice the array (length of signature) - possible signature
                    if possible signature == signature:
                        counter += 1

        print(counter)

Python 3代码：

#!/usr/bin/python
# -*- coding: utf-8 -*-
noc = int(input(''))  # getting the number of cases # NOC- number of cases

# Loop over the # of cases

for c in range(noc):
    (N, K) = [int(i) for i in input('').split(' ')]  # getting N, K

    # N - number of elements given
    # K - initial countdown value
    # getting the elements

    caseList = [int(i) for i in input('').split(' ')]

    # generating a 'signature' or list of factorial for the countdown

    steps = [i for i in range(1, K + 1)][::-1]

    # counter for number of matches

    countdown = 0  # init value

    # loop over each element i  n list

    for i in range(N):

        # if element == K(init countdown number)

        if caseList[i] == K:

            # make sure there is space to get the sliced array

            if i <= len(caseList) - len(steps):

                # get the next m numbers if

                if caseList[i:i + len(steps)] == steps:
                    countdown += 1  # increment
    print countdown  # print the number of matches

Answer 1

您的解决方案看起来不错，除了输出不是指定的，不是针对Python 3，而是针对2，只需将其更改为：

print(f'Case {c}: {countdown}')

除此之外，您要做的工作比需要的多。您实际上只需要遍历整个列表一次即可计算K倒计时。

例如：

import sys
from io import StringIO

sys.stdin = StringIO('3\n2 2\n2 1\n8 2\n0 2 1 2 2 1 2 1 0\n0 2\n\n')

t = int(input())
for c in range(t):
    (n, k) = [int(i) for i in input().split()]
    a = [int(i) for i in input().split()]
    # initialise goal, position in array and count
    goal, i, count = k, 0, 0
    while i < n:
        # if item in current position matches current goal
        if a[i] == goal:
            # then next goal item is one less
            goal -= 1
            # until all in K-countdown were found
            if goal == 0:
                # then start over and increase count
                goal = k
                count += 1
            # look at the next position
            i += 1
        # else (current item doesn't match goal), if already looking for start of K-countdown
        elif goal == k:
            # look at the next position
            i += 1
        # else (current item doesn't match goal, goal wasn't start of K-countdown)
        else:
            # look for start of K-countdown
            goal = k
    print(f'Case #{c}: {count}')

Answer 2

我发现您的解决方案没有任何问题。可能是您的输出格式。

您应该以Case #x的形式输出：y，其中x是测试用例编号（从1开始），y是她数组中K倒数的数量。

示例：

案例1：2

案例2：0

案例3：1

注意：如果您使用的是print x而不是print(x)，请确保使用的是Python 2.x

Answer 3

我也在想同样的事情。

让我们看看给定的约束：

1 ≤ T ≤ 100. # Testcases
2 ≤ K ≤ N. # Value of K
1 ≤ Ai ≤ 2 × 105, for all i. # Index- i

# Test Set 1
2 ≤ N ≤ 1000.

# Test Set 2
2 ≤ N ≤ 2 × 105 for at most 10 test cases.
For the remaining cases, 2 ≤ N ≤ 1000.

现在假设我们有一个测试用例

nums = [1]
k = 1

有人可能会认为K = 1倒数= 1对吗？其实不是。

仔细阅读2 <= N，表示

数组长度必须为最小长度= 2。

Expected result,
nums = [1]
K    = 1
coutdown = 0

Answer 4

当约束已经说 2<=N 是不是就没有数组长度=0或1的测试用例了

Answer 5

@MFK34 没有问题，除了 print() 在 python 3 中需要括号，他在循环结束时立即打印答案并且解决方案不符合预期。以下是我修改后的解决方案。

#!/usr/bin/python
# -*- coding: utf-8 -*-
noc = int(input(''))  # getting the number of cases # NOC- number of cases
op = []

# Loop over the # of cases

for c in range(noc):
    (N, K) = [int(i) for i in input('').split(' ')]  # getting N, K

    caseList = [int(i) for i in input('').split(' ')]

    steps = [i for i in range(1, K + 1)][::-1]

    # counter for number of matches

    countdown = 0  # init value
    # loop over each element i  n list

    for i in range(N):

    # if element == K(init countdown number)

    if caseList[i] == K:

        # make sure there is space to get the sliced array

        if i <= len(caseList) - len(steps):

            # get the next m numbers if

            if caseList[i:i + len(steps)] == steps:
                countdown += 1  # increment
op.append(countdown)

for i,d in enumerate(op):
    print("Case #"+str(i+1)+":",d)

我刚刚将结果存储在一个数组中，然后按预期顺序打印在输入的末尾。

Google Kickstart 2020轮C-倒计时。无法理解为什么我的解决方案不正确

问题

伪代码

Python 3代码：

5 个答案: