Given an integer array?nums?and an integer?k, return?the number of non-empty?subarrays?that have a sum divisible by?k.

A?subarray?is a?contiguous?part of an array.

?

Example 1:

Input: nums = [4,5,0,-2,-3,1], k = 5Output: 7Explanation: There are 7 subarrays with a sum divisible by k = 5: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Example 2:

Input: nums = [5], k = 9Output: 0

?

Constraints:

• 1 <= nums.length <= 3 * 104

• -104 <= nums[i] <= 104

• 2 <= k <= 104

利用前綴和，求出余數(shù)，然后把余數(shù)放到map 中，后續(xù)根據(jù)map的value，任取2個(gè)相同的key的對(duì)應(yīng)值，也就是C(n,2)-(n*(n-1)/2)了，但是不能忽略余數(shù)為0的值，需要單獨(dú)加一次、

其他的直接取2個(gè)就可以，但是余數(shù)為0的，自己就能可以的，所以要加上；

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