Continuous Subarray Sum

本次題目是有關於動態規劃的應用，Continuous Subarray Sum，題目的大致要求如下:

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

對於求解的過程，我的思路大致如下，首先考慮給定數列n，並構造其各階級數

S(n) = a(1) + a(2) + ... + a(n), n ≥ 1，

進一步可以考慮遞歸算法

S(n) = a(1), n = 1

S(n) = a(n) + S(n - 1), n > 1，

考慮到子數列的和可以寫成這樣的形式

S(n) = a(1), n = 1

S(n) = a(n) + S(n - 1), n > 1。

有了這樣的構思就方便我們去設計此代碼

bool checkSubarraySum(vector<int>& nums, int k)
{
map<int, int> sum; //構造各階級數
for(int i = 0; i < nums.size(); i++)
{
sum[i] = sum[i - 1] + nums[i];
} 

for(int i = 0; i < nums.size(); i++)
{
for(int j = i + 1; j < nums.size(); j++)
{
if (k != 0 && (sum[j] - sum[i - 1]) % k == 0 || k == 0 && (sum[j] - sum[i - 1]) == 0) 
{
return true;
}
}
}
return false;
}

應用到動態規劃的概念，if (k != 0 && (sum[j] - sum[i - 1]) % k == 0 || k == 0 && (sum[j] - sum[i - 1]) == 0) 判斷每一種不同子數列相加或n*k(mode k)的可能。