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.

1 8

1123123111

## 解答

#include<iostream>
#include<string>

using namespace std;

int main(){
string str;
int n, j = 0;
cin >> str >> n;
for (int k = 1; k < n; k++) {
string tmp;
for (int i = 0; i < str.length(); i = j) {
for (j = i; j < str.length() && str[j] == str[i]; j++);
tmp += str[i] + to_string(j - i);
}
str = tmp;
}
cout << str;
return 0;
}