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
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 int main(){ 5 cin >> n >> m; 6 string s, st; 7 s += n+'0'; 8 m--; 9 while(m--){ 10 char c = s[0]; 11 int num = 1; 12 for(int i = 1; i < s.length(); i++){ 13 if(c == s[i]){ 14 num++; 15 }else{ 16 st += c; 17 st += '0'+num; 18 c = s[i]; 19 num = 1; 20 } 21 } 22 st += c; 23 st += '0'+num; 24 s = st; 25 st = ""; 26 } 27 cout << s <<endl; 28 return 0; 29 }