python 整除的数组_计算和可被整除的所有子数组

在我学习面试的时候，我在GeeksForGeeks上找到了这个问题和解决方案，但不明白答案。在

上面说的是Let there be a subarray (i, j) whose sum is divisible by k

sum(i, j) = sum(0, j) - sum(0, i-1)Sum for any subarray can be written as q*k + rem where q is a

quotient and rem is remainder Thus,

^{pr2}$

We see, for sum(i, j) i.e. for sum of any subarray to be

divisible by k, the RHS should also be divisible by k.

(q1 - q2)k is obviously divisible by k, for (rem1-rem2) to

follow the same, rem1 = rem2 where

^{3}$

首先，我不知道q1和{}表示什么。在def subCount(arr, n, k):

# create auxiliary hash

# array to count frequency

# of remainders

mod =[]

for i in range(k + 1):

mod.append(0)

cumSum = 0

for i in range(n):

cumSum = cumSum + arr[i]

mod[((cumSum % k)+k)% k]= mod[((cumSum % k)+k)% k] + 1

result = 0 # Initialize result

# Traverse mod[]

for i in range(k):

if (mod[i] > 1):

result = result + (mod[i]*(mod[i]-1))//2

result = result + mod[0]

return result

在这个解决方案代码中，我没有得到mod的角色。增加第((cumSum % k)+k)% k个数组的数目有什么效果？在

如果能一步一步地解释清楚那就太好了。谢谢。在