Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, ...

whereDis 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 oneDin the 1st number, and hence it isD1; the 2nd number consists of oneD

(corresponding toD1) and one 1 (corresponding to 11), therefore the 3rd number isD111; or since the 4th number isD113, it consists of oneD, two 1's, and one 3, so the next number must beD11231. This definition works forD= 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digitD.

Input Specification:

Each input file contains one test case, which givesD(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 ofD.

Sample Input:

1 8

Sample Output:

1123123111

題解：遍歷字串，每遇到一個字元就下，並迴圈比較後一位是否與之相同，設立計數器計數

#include<bits/stdc++.h>
using
 
 namespace std;
int main(){
    string s;
    int n;
    cin>>s>>n;
    for(int i=0;i<n-1;i++){
        string ss="";
        int cnt=0;
        for(int j=0;j<s.length();j++){
            ss+=s[j];
            cnt=0;
            while(s[j+1]==s[j]){
                cnt++;
                j++;
            }
            if(cnt>0){
                cnt++;
                ss+=(cnt+'0');
            }
            else{
                ss+='1';
            }
        }
        s=ss;
    }
    cout<<s<<endl;
    return 0;
}