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 8Sample Output:
1123123111#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; }