1140 Look-and-say Sequence_look-and-say sequence is a sequence of integers as-优快云博客

本文链接：https://blog.youkuaiyun.com/ssf_cxdm/article/details/82951295

本文详细介绍了一种基于描述字符串的看说数列算法，通过具体实例解释了算法的工作原理，包括输入数字序列的生成规则及如何通过递归方式生成后续数列。文章提供了完整的C++实现代码，帮助读者理解并实现该算法。

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1140 Look-and-say Sequence
题目链接
Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, …
where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1’s, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.

Input Specification:
Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (≤ 40), separated by a space.

Output Specification:
Print in a line the Nth number in a look-and-say sequence of D.

Sample Input:
1 8
Sample Output:
1123123111
题目大意：对字符串进行描述
例如1123123111，这样描述2个1，1个2，1个3，1个1，1个2，1个3，3个1。即12213111213113
参考代码：

#include<iostream>
#include<string>
using namespace std;
string func(string s) {
	string res;
	for (int i = 0; i < s.size(); i++) {
		int j = i + 1, cnt = 1;
		res = res + to_string(s[i] - '0');
		//res = res + s[i];超时
		while (j < s.size()&&s[j]==s[i]){
			j++;
			cnt++;
		}
		res = res + to_string(cnt);
		i += (cnt - 1);
	}
	return res;
}
int main() {
	int D, N, cnt = 1;
	cin >> D >> N;
	string s = to_string(D);
	while (cnt < N) {
		s = func(s);
		cnt++;
	}
	cout << s;

	return 0;
}