[pull] master from wisdompeak:master #339

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		[2999.Count-the-Number-of-Powerful-Integers](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/2999.Count-the-Number-of-Powerful-Integers) (H-)
		[3307.Find-the-K-th-Character-in-String-Game-II](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/3307.Find-the-K-th-Character-in-String-Game-II) (M)
		[3490.Count-Beautiful-Numbers](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/3490.Count-Beautiful-Numbers) (M+)
	[3614.Process-String-with-Special-Operations-II](https://github.com/wisdompeak/LeetCode/tree/master/Recursion/3614.Process-String-with-Special-Operations-II) (H-)

		#### [Graph](https://github.com/wisdompeak/LeetCode/tree/master/Graph/)
		[332.Reconstruct-Itinerary](https://github.com/wisdompeak/LeetCode/tree/master/DFS/332.Reconstruct-Itinerary) (H)
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	using ll = long long;
	class Solution {
	public:
	char processStr(string s, long long k) {
	k++;
	int n = s.size();
	s = "#"+s;
	vector<ll>len(n+1);
	const ll INF = 1e15+5;

	for (int i=1; i<=n; i++) {
	char c = s[i];
	ll prev = len[i-1];
	ll now = prev;
	if ('a'<=c && c<='z')
	now = prev+1;
	else if (c == '*')
	now = prev>0 ? (prev-1): 0;
	else if (c=='#')
	now = prev*2;
	else if (c=='%')
	now = prev;
	len[i] = min(now, INF);
	}
	if (k==0 \|\| k>(ll)len[n]) return '.';

	for (int i=n; i>=1; i--) {
	char c = s[i];
	ll before = len[i-1];
	ll after = len[i];
	if ('a'<=c && c<='z') {
	if (k==after)
	return c;
	} else if (c=='*') {

	} else if (c=='#') {
	if (k>before)
	k-=before;
	} else if ( c=='%') {
	k = before-k+1;
	}
	}

	return '.';
	}
	};

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	### 3614.Process-String-with-Special-Operations-II

	很显然,构造完全之后的字符串非常长,我们不可能在此基础上定位第k个元素。此题极大概率是递归反推。

	我们分类思考每种操作下的第k个字符(注意是1-index)是什么情况:
	1. 添加一个字符,使得长度从a变成了a+1. 如果k=a+1,那么答案就是最新添加的字符。否则,递归求原字符串里的第k个。
	2. 删减最后一个字符,使得长度从a变成了a-1. 这种情况下k不可能是a(否则无解),因此等效于递归求原字符串里的第k个。
	3. 将字符串copy+append,使得长度从a变成了2a. 显然如果k<=a,递归求原字符串里的第k个。如果k>a,递归求原字符串里的第k-a个。
	4. 将字符串反转,长度依然是k。显然,递归求原字符串里的第a+1-k个。

	综上,只要知道每个回合的操作和长度变化,我们就可以把"求当前字符串的第k个元素",逆推为"求前一个回合字符串的第k'个元素",其中k'的计算如上。

	注意,这个题可能无解。逆推的前提是正向的变化合法存在。所以我们必须先正向走一遍,记录下每个回合的字符串长度,最后检查k是否在最终长度的范围内。

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