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

思路：

考試的時候卡這道題卡了很久，終究是沒讀懂題...

第n+1個串是第n個串的描述。   比如第n個串是11122234（隨便寫的），那麼第n+1個串就是13233141，這個意思就是在第n個串中  1有3個， 2有3個， 3有1個， 4有1個， 所以是13233141。 

C++：

#include "cstdio"
#include "iostream"
#include "string"
using namespace std;
int main(){
	int n,k;
	string s,ans;
	cin>>s>>n;
	for (int i=1;i<n;i++)
	{
		string t;
		for (int j=0;j<s.length();j=k)
		{
			for (k=j;k<s.length()&&s[k]==s[j];k++);
			t+=to_string(s[j]-'0')+to_string(k-j);
		}
		s=t;
	}
	cout<<s<<endl;
	return 0;
}