You are given a?0-indexed?integer array?nums?of length?n.

You can perform the following operation as many times as you want:

• Pick an index?i?that you haven’t picked before, and pick a prime?p?strictly less than?nums[i], then subtract?p?from?nums[i].

Return?true if you can make?nums?a strictly increasing array using the above operation and false otherwise.

A?strictly increasing array?is an array whose each element is strictly greater than its preceding element.

?

Example 1:

Input: nums = [4,9,6,10]

Output: true

Explanation:?

In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0],?

so that nums becomes [1,9,6,10].?

In the second operation: i = 1, p = 7, subtract 7 from nums[1],?

so nums becomes equal to [1,2,6,10].?

After the second operation, nums is sorted in strictly increasing order, so the answer is true.

Example 2:

Input: nums = [6,8,11,12]

Output: true

Explanation: Initially nums is sorted in strictly increasing order, so we don't need to make any operations.

Example 3:

Input: nums = [5,8,3]

Output: false

Explanation: It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.

?

Constraints:

• 1 <= nums.length <= 1000

• 1 <= nums[i] <= 1000

• nums.length == n

好尷尬啊，基本上每個(gè)函數(shù)都寫錯(cuò)過，isprime函數(shù)漏寫了1的情況，1不是質(zhì)數(shù)，不是質(zhì)數(shù)，不是質(zhì)數(shù)。。。。

first_numSubtract()這個(gè)函數(shù)是將第一個(gè)元素先降低到最小值，

然后剩下的元素去每次判斷是否存在num[i]-num[i-1]的質(zhì)數(shù)，有的話，就將這個(gè)num[i]減去對應(yīng)的質(zhì)數(shù)，

還有個(gè)checksort的函數(shù)，判斷是否是嚴(yán)格遞增的，沒錯(cuò)是嚴(yán)格遞增，我以為>=就行了，不行，不行，不行，要>才行啊。。。

綜上就是題意沒理解透徹，另外就是基礎(chǔ)不扎實(shí)，還需要多多鍛煉，汗顏啊。。。

Runtime:?6 ms, faster than?90.91%?of?Java?online submissions for?Prime Subtraction Operation.

Memory Usage:?42.6 MB, less than?63.64%?of?Java?online submissions for?Prime Subtraction Operation.