1140 Look-and-say Sequence （20 分）

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

题意：后面一个数是前面一个数的描述，一般第一个数是d，代表0-9的任意一个数，第二 数是第一个数的描述，就是将d+d的个数。同样，第三个数是第二个数的描述，依次，例如:1   11(前一个1是第一个数，后一个1是第一个中1的个数)    12（代表前一个数中有2个1）  1121（前面一个数中有1个1，1个2，数放前，个数放后）  122111   112213  12221131  1123123111 。

#include
#include
using namespace std;
int main(){
	int n;
	string s; 
	cin >> s >> n;
	for(int k = 2;k <= n;k++){
		string t; 
		int top = 1;
		for(int j = 1;j < s.length()+1;j++){
			if(s[j] == s[j-1]){
				top++;
			}else{
				t+=s[j-1]+to_string(top);
				top = 1;
			}
		}
		s = t;
		//cout << t << endl;
	}
	cout << s;
	return 0;
}