You are given an integer array nums and three integers k, op1, and op2.
You can perform the following operations on nums:
i and divide nums[i] by 2, rounding up to the nearest whole number. You can perform this operation at most op1 times, and not more than once per index.i and subtract k from nums[i], but only if nums[i] is greater than or equal to k. You can perform this operation at most op2 times, and not more than once per index.Note: Both operations can be applied to the same index, but at most once each.
\n\nReturn the minimum possible sum of all elements in nums after performing any number of operations.
\n
Example 1:
\n\nInput: nums = [2,8,3,19,3], k = 3, op1 = 1, op2 = 1
\n\nOutput: 23
\n\nExplanation:
\n\nnums[1] = 8, making nums[1] = 5.nums[3] = 19, making nums[3] = 10.[2, 5, 3, 10, 3], which has the minimum possible sum of 23 after applying the operations.Example 2:
\n\nInput: nums = [2,4,3], k = 3, op1 = 2, op2 = 1
\n\nOutput: 3
\n\nExplanation:
\n\nnums[0] = 2, making nums[0] = 1.nums[1] = 4, making nums[1] = 2.nums[2] = 3, making nums[2] = 0.[1, 2, 0], which has the minimum possible sum of 3 after applying the operations.\n
Constraints:
\n\n1 <= nums.length <= 1000 <= nums[i] <= 1050 <= k <= 1050 <= op1, op2 <= nums.length给你一个整数数组 nums 和三个整数 k、op1 和 op2。
你可以对 nums 执行以下操作：
i，将 nums[i] 除以 2，并 向上取整 到最接近的整数。你最多可以执行此操作 op1 次，并且每个下标最多只能执行一次。i，仅当 nums[i] 大于或等于 k 时，从 nums[i] 中减去 k。你最多可以执行此操作 op2 次，并且每个下标最多只能执行一次。注意： 两种操作可以应用于同一下标，但每种操作最多只能应用一次。
\n\n返回在执行任意次数的操作后，nums 中所有元素的 最小 可能 和 。
\n\n
示例 1：
\n\n输入： nums = [2,8,3,19,3], k = 3, op1 = 1, op2 = 1
\n\n输出： 23
\n\n解释：
\n\nnums[1] = 8 应用操作 2，使 nums[1] = 5。nums[3] = 19 应用操作 1，使 nums[3] = 10。[2, 5, 3, 10, 3]，在应用操作后具有最小可能和 23。示例 2：
\n\n输入： nums = [2,4,3], k = 3, op1 = 2, op2 = 1
\n\n输出： 3
\n\n解释：
\n\nnums[0] = 2 应用操作 1，使 nums[0] = 1。nums[1] = 4 应用操作 1，使 nums[1] = 2。nums[2] = 3 应用操作 2，使 nums[2] = 0。[1, 2, 0]，在应用操作后具有最小可能和 3。\n\n
提示：
\n\n1 <= nums.length <= 1000 <= nums[i] <= 1050 <= k <= 1050 <= op1, op2 <= nums.lengthdp[index][op1][op2] where each state tracks progress at index with op1 and op2 operations left.",
"At each state, try applying only operation 1, only operation 2, both in sequence, or skip both to find optimal results."
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