1. 題目

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. 題意

外觀數列，如下案例：

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

其中D是一個[0, 9]範圍內不等於1的整數；

數列第2項：表示第1項有1個D，所以為D1；

數列第3項：表示第2項中有1個1和1個D，所以為D111；

數列第4項：表示第3項中有1個D和3個1，所以為D113；

數列第5項：表示第4項中有1個D，2個1和1個3，所以為D11231；

...

數列第n項：...

題目要求求解給定數字D的外觀數列的第n項。

3. 思路

外觀數列變換： 根據題意，每次新的外觀數列是對上一次外觀數列的一次變換。

變化規則：即對上一個外觀數列字串中的連續字元進行計數，並將字元和字元個數合併成新的字串即為新的外觀數列。

4. 程式碼

#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)
		{
			// 計數連續相等字元的個數 
			cnt++;
		} else
		{
			// 當出現字元不一致時，計數結束，並將字元和個數加入結果temp字串 
			temp += ch + to_string(cnt);
			ch = res[i];
			cnt = 1;
		}
	}
	// 字串最後一位沒有進行計數操作，額外進行一次操作 
	temp += ch + to_string(cnt);
	// 將最新外觀數列temp賦值給res 
	res = temp;
}

int main()
{
	int n;
	cin >> res >> n;
	// 外觀數列的第一項即為本身，不需要進行額外計算 
	n -= 1;	
	while (n--)
	{
		lookAndSay();
	}
	cout << res << endl;
	return 0;
}