1. 题目

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

2. 题意

外观数列，如下案例：

D, D1, D111, D113, D11231...

其中D是一个[0, 9]范围内不等于1的整数；

数列第2项：表示第1项有1个D，所以为D1；

数列第3项：表示第2项中有1个1和1个D，所以为D111；

数列第4项：表示第3项中有1个D和3个1，所以为D113；

数列第5项：表示第4项中有1个D，2个1和1个3，所以为D11231；

...

数列第n项：...

题目要求求解给定数字D的外观数列的第n项。

3. 思路

外观数列变换： 根据题意，每次新的外观数列是对上一次外观数列的一次变换。

变化规则：即对上一个外观数列字符串中的连续字符进行计数，并将字符和字符个数合并成新的字符串即为新的外观数列。

4. 代码

#include <iostream>
#include <string>

using namespace std;

string res;

void lookAndSay()
{
	char ch = res[0];
	string temp = "";
	int cnt = 1;
	for (int i = 1; i < res.length(); ++i)
	{
		if (res[i] == ch)
		{
			// 计数连续相等字符的个数 
			cnt++;
		} else
		{
			// 当出现字符不一致时，计数结束，并将字符和个数加入结果temp字符串 
			temp += ch + to_string(cnt);
			ch = res[i];
			cnt = 1;
		}
	}
	// 字符串最后一位没有进行计数操作，额外进行一次操作 
	temp += ch + to_string(cnt);
	// 将最新外观数列temp赋值给res 
	res = temp;
}

int main()
{
	int n;
	cin >> res >> n;
	// 外观数列的第一项即为本身，不需要进行额外计算 
	n -= 1;	
	while (n--)
	{
		lookAndSay();
	}
	cout << res << endl;
	return 0;
}