See till count<=m it is clear, that we have to eat minimum possible sweet(s) that day. But after that, what we got to do is minimize the sum, which means that larger sweets(Ai) have to be consumed earlier. So for every count>m, we will eat the largest m sweets that day, but the remaining smaller ones next day, and the even smaller ones the day after. Now see that the ans array will have the prefix sum at each index, so when its day 2, we just need to add the minimum of remaining sweets(k-m) to the current sum, and that's why we are doing ans[i]+=ans[i-m]. Its best if you perform a simple check with pen and paper using the sample input. I am sure you will figure it out.

In this problem, it is always optimal to eat the m sweetest sweets first if you have to eat them so they do not get multiplied.

If i <= m, the answer is just the prefix sum because Yui can just eat all the candies on the first day.

sum(a, b) refers to the sum of the interval of a to b inclusive of candies in ascending order.

To find dp[i] if i > m, Yui has to do the following:

1. Eat the m sweetest candies on the first day: sum(i-m+1, i)

2. Eat the i-m candies left and increase that value by a factor of 1 because it would take Yui an additional day to eat these sweets: dp[i-m] + sum(1, i-m)

So the complete result is

sum(i-m+1, i) + dp[i-m] + sum(1, i-m)

=> dp[i-m] + sum(1, i)

=> dp[i-m] + (prefix sum up to i inclusive)

Hope this helps!