一个人将数字列表交给 String 先生，但 String 先生只懂字符串。在字符串，他也只能理解元音。 String 先生需要您的帮助来找出加起来等于某个数字D的对的总数。
数字D的计算规则如下：
- 获取所有数字并将它们转换为它们的文本表示。
- 接下来，从所有文本表示中总结元音的数量，即{a, e, i, o, u} 。这个总和是数字D 。
- 现在，一旦知道数字D ，找出输入中所有和等于D 的无序数字对。
问题陈述：给定一个由N ( 1 ≤ N ≤ 100 ) 个整数组成的数组arr[] ，任务是将每个数组元素 ( 1 ≤ arr[i] ≤ 100 ) 转换为它们各自的文本表示并打印数组中所有可能对的计数，其总和等于其文本表示中存在的元音总数。如果计数超过 100，则打印“greater 100” 。
注意：对于数字 100，将其转换为文本表示形式为百而不是百。
例子：
Input: arr[] = {1, 2, 3, 4, 5}
Output: one
Explanation:
1 -> one -> o, e (2 vowels)
2 -> two -> o (1 vowel)
3 -> three -> e, e (2 vowels)
4 -> four -> o, u (2 vowels)
5 -> five – > i, e (2 vowels)
The total count of vowels in their textual representations = {2 + 1 + 2 + 2 + 2} = 9.
Now from the given array, only a single unordered pair {4, 5} sums up to 9. Therefore, the count is 1. Hence, the required output is “one“.
Input: arr[] = {7, 4, 2, }
Output: zero
Explanation:
7 -> seven -> e, e (2 vowels)
4 -> four -> o, u (2 vowels)
2 -> two -> o (1 vowel)
The total count of vowels in their textual representation = {2 + 2 + 1} = 5.
Now from the given array, no pair exists which adds up to 5. Therefore, the answer is “zero“.
方法：按照以下步骤解决此问题：
- 在Map 中存储从0到100的每个数字的文本表示。
- 遍历数组，对于每个数组元素，将每个数字转换为其文本形式。
- 找出文本形式中存在的元音总数并将其存储在一个变量中，比如D 。
- 现在，从给定的数组中生成所有可能的对。
- 计算总和为D 的所有对。
- 如果计数超过 100，则“大于 100” 。否则，打印其文本形式。
下面是上述方法的实现：
Python3
# Python3 program for the above approach
# Import combinations
from itertools import combinations
# Function to check the string pair
def string_pair(n, nums):
words = {0: 'zero', 1: 'one', 2: 'two', 3: 'three',
4: 'four', 5: 'five', 6: 'six', 7: 'seven',
8: 'eight', 9: 'nine', 10: 'ten', 11: 'eleven',
12: 'twelve', 13: 'thirteen', 14: 'fourteen',
15: 'fifteen', 16: 'sixteen', 17: 'seventeen',
18: 'eighteen', 19: 'nineteen', 20: 'twenty',
21: 'twentyone', 22: 'twentytwo', 23: 'twentythree',
24: 'twentyfour', 25: 'twentyfive', 26: 'twentysix',
27: 'twentyseven', 28: 'twentyeight', 29: 'twentynine',
30: 'thirty', 31: 'thirtyone', 32: 'thirtytwo',
33: 'thirtythree', 34: 'thirtyfour', 35: 'thirtyfive',
36: 'thirtysix', 37: 'thirtyseven', 38: 'thirtyeight',
39: 'thirtynine', 40: 'forty', 41: 'fortyone',
42: 'fortytwo', 43: 'fortythree', 44: 'fortyfour',
45: 'fortyfive', 46: 'fortysix', 47: 'fortyseven',
48: 'fortyeight', 49: 'fortynine', 50: 'fifty',
51: 'fiftyone', 52: 'fiftytwo', 53: 'fiftythree',
54: 'fiftyfour', 55: 'fiftyfive', 56: 'fiftysix',
57: 'fiftyseven', 58: 'fiftyeight', 59: 'fiftynine',
60: 'sixty', 61: 'sixtyone', 62: 'sixtytwo',
63: 'sixtythree', 64: 'sixtyfour', 65: 'sixtyfive',
66: 'sixtysix', 67: 'sixtyseven', 68: 'sixtyeight',
69: 'sixtynine', 70: 'seventy', 71: 'seventyone',
72: 'seventytwo', 73: 'seventythree', 74: 'seventyfour',
75: 'seventyfive', 76: 'seventysix', 77: 'seventyseven',
78: 'seventyeight', 79: 'seventynine', 80: 'eighty',
81: 'eightyone', 82: 'eightytwo', 83: 'eightythree',
84: 'eightyfour', 85: 'eightyfive', 86: 'eightysix',
87: 'eightyseven', 88: 'eightyeight', 89: 'eightynine',
90: 'ninety', 91: 'ninetyone', 92: 'ninetytwo',
93: 'ninetythree', 94: 'ninetyfour', 95: 'ninetyfive',
96: 'ninetysix', 97: 'ninetyseven', 98: 'ninetyeight',
99: 'ninetynine', 100: 'hundred'}
# Map the string into list of integers
nums = list(map(int, nums))
# Temporary lists to store list of count
ls, ls1 = [], []
count, c = 0, 0
# Iterating through the numbers
for i in nums:
# Stores the textual form of i
s = words[i]
for a in range(len(s)):
vo = ['a', 'e', 'i', 'o', 'u']
# If it is vowel
if s[a] in vo:
# Increment the count
count += 1
# Append the count
ls.append(count)
count = 0
# D = sum(count of vowels)
d = sum(ls)
for i in nums:
# To find the numbers less
# that or equal to d,
# so as to find the pair sum
if i <= d:
# Append the numbers
# whose sum can be d
ls1.append(i)
# Stores all possible pairs in the
# form of an object list of tuples
comb = combinations(ls1, 2)
# Traverse all the pairs
for i in list(comb):
# If sum is equal to d
if sum(i) == d:
# Increment count
c += 1
# If count exceeds 100
if c <= 100:
print(words)
# Otherwise
else:
print("greater 100")
# Driver Code
if __name__ == '__main__':
# Given Length of string
n = 5
# Given array
arr = [1, 2, 3, 4, 5]
# Function Call
string_pair(n, arr)one
时间复杂度： O(N 2 )
辅助空间： O(N 2 )
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