1. Complexity [20 marks] (A)Prove that is both and . [4 marks] (B)Consider the following functions of . Put them in order from smallest to largest asymptotic growth rate. [8 marks] (C)Let be the processing time of an algorithm for the input of size . Which is the asymptotic time complexity of this algorithm, or ? Please show your working to justify your answer. [8 marks]
时间: 2023-03-12 13:04:39 浏览: 74
A:证明 既是 又是 。[4 分]
B:考虑 的以下函数。将它们按从最小到最大的渐近增长率排序。[8 分]
C:让 是算法对输入大小 的处理时间。该算法的渐近时间复杂度是 还是 ?请出示您的计算步骤以证明您的答案。[8 分]
相关问题
You are given two arrays a and b , both of length n . Your task is to count the number of pairs of integers (i,j) such that 1≤i<j≤n and ai⋅aj=bi+bj .
To solve this problem, we can first preprocess the array b to create a hash table that maps each value in b to its frequency. Then, for each pair of indices (i,j) such that i<j, we can check if ai⋅aj is equal to bi+bj by computing ai/bj and aj/bi and checking if they are equal. If they are equal, we can check if the values ai/bj and aj/bi are present in the hash table. If both values are present, we can increment our count by the product of their frequencies in the hash table.
The time complexity of this algorithm is O(n^2) due to the nested loops, but the constant factor is small due to the hash table lookups. Here is the Python code for the algorithm:
```
from collections import defaultdict
def count_pairs(a, b):
n = len(a)
freq = defaultdict(int)
for x in b:
freq[x] += 1
count = 0
for i in range(n):
for j in range(i+1, n):
if a[i]*a[j] == b[i]+b[j]:
if b[i] != 0 and b[j] % b[i] == 0 and a[j] == (b[j] // b[i]) * a[i]:
x = b[i]
y = b[j] // b[i]
if x in freq and y in freq:
count += freq[x] * freq[y]
if b[j] != 0 and b[i] % b[j] == 0 and a[i] == (b[i] // b[j]) * a[j]:
x = b[j]
y = b[i] // b[j]
if x in freq and y in freq:
count += freq[x] * freq[y]
return count
```
Note that we need to check both possible orderings of ai and aj in case one of them is zero or the corresponding value in b is not present in the hash table.
You are given two arrays a and b,both of length n Your task is to count the number of pairs of integers (i,j) such that 1≤i<j≤n and ai⋅aj=bi+bj .
To count the number of pairs of integers (i,j) such that ai⋅aj=bi+bj, we can use a hash table to keep track of the count of each value in array a and b. Then, we can loop through all pairs of indices (i,j) and check if ai⋅aj=bi+bj. If it is true, we increment a counter.
Here is the code:
```
def count_pairs(a, b):
n = len(a)
count_a = {}
count_b = {}
for i in range(n):
count_a[a[i]] = count_a.get(a[i], 0) + 1
count_b[b[i]] = count_b.get(b[i], 0) + 1
count = 0
for i in range(n):
for j in range(i+1, n):
if a[i]*a[j] == b[i]+b[j]:
count += 1
return count
```
Note that this solution has a time complexity of O(n^2) because of the nested loops. However, we can make it more efficient by using a set to store all possible values of ai⋅aj and bi+bj, and then check if each product is in the set. This reduces the time complexity to O(n).