Given an array A
of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K
.
Example 1:
Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: 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]
Note:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
给定一个整数数组
A
,返回其中元素之和可被 K
整除的(连续、非空)子数组的数目。
示例:
输入:A = [4,5,0,-2,-3,1], K = 5 输出:7 解释: 有 7 个子数组满足其元素之和可被 K = 5 整除: [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
提示:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
368ms
1 class Solution { 2 func subarraysDivByK(_ A: [Int], _ K: Int) -> Int { 3 var n = A.count 4 var f:[Int] = [Int](repeating:0,count:K) 5 f[0] = 1 6 var x:Int = 0 7 for v in A 8 { 9 x += v 10 x %= K 11 if x < 0 12 { 13 x += K 14 } 15 f[x] += 1 16 } 17 var ret:Int = 0 18 for i in 0..<K 19 { 20 ret += f[i]*(f[i]-1)/2 21 } 22 return ret 23 } 24 }
原文地址:https://www.cnblogs.com/strengthen/p/10262262.html
时间: 2024-10-09 03:40:30