最大配对乘积的Python

本文介绍了最大配对乘积的Python的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 n = int(input("Enter the number of elements in the array (2-200,000):")) a = [int(x) for x in input("Enter all numbers of the sequence with only non-negative intergers not exceeding 100,000:").split()] c = list()3 for i in range(0,n): for j in range (1,n): if a[i] != a[j]: m = a[i]*a[j] c.append(m) else: continue print(max(c))

此代码可以工作。但是，我想定义一个函数来自动计算以下代码中第5行的最大乘积。

def MaxPairwiseProduct(n,a,c): for i in range(0,n): for j in range (1,n): if a[i] != a[j]: m = a[i]*a[j] c.append(m) else: continue Product = max(c) return Product n = int(input("Enter the number of elements in the array (2-200,000):")) a = [int(x) for x in input("Enter all numbers of the sequence with only non-negative intergers not exceeding 100,000:").split()] c = list() MaxPairwiseProduct(n,a,c)

我重写了该函数，但它不起作用。它显示"IndentationError：预期为缩进块"

推荐答案# Uses python3 def MaxPairwiseProduct(n,a,c): for i in range(0,n): for j in range (1,n): if a[i] != a[j]: m = a[i]*a[j] c.append(m) else: continue Product1 = max(c) return Product1 def MaxPairwiseProductFast(n,a): max_index1 = -1 for i in range(0,n): if a[i] > a[max_index1]: max_index1 = i else: continue #the value of the other index should be different compared to the #first, else it will assume the same indices for both the max max_index2 = -2 for j in range(0,n): if a[j] > a[max_index2] and a[j] != a[max_index1]: max_index2 = j else: continue Product2 = a[max_index1]*a[max_index2] return Product2 n = int(input("Enter the number of elements in the array (2-200,000):")) a = [int(x) for x in input("Enter all numbers of the sequence with only non-negative intergers not exceeding 100,000:").split()] c = list() print('The max value by regular algorithm:', MaxPairwiseProduct(n,a,c)) print('The max value by faster algorithm:', MaxPairwiseProductFast(n,a))

此代码包含计算最大值的第二种算法。