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
#include<iostream>
#include<vector>
#include<cstdlib>
using namespace std;
int main(){
int d, n;
cin >> d >> n;
string s = to_string(d);
for(int j = 0; j < n - 1; j++){
string temp_s = "";
char c = s[0];
int count = 1;
for(int i = 1; i < s.size(); i++){
if(c == s[i]){
count++;
}else
{
// cout << c << " " << count << endl;
temp_s += c;
temp_s += to_string(count);
count = 1;
c = s[i];
}
}
// cout << c << " " << count << endl;
temp_s += c;
temp_s += to_string(count);
s = temp_s;
}
cout << s << endl;
return 0;
}