#1140. Look-and-say Sequence【模拟】

原题链接

Problem Description：

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 $N$ ($\le 40$), 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

Problem Analysis:

注意第 $n+1$ 项统计的是第 $n$ 项每个字符连续出现的个数。

Code

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
    
    
    int n, d;
    cin >> d >> n;

    string cur = to_string(d);
    for (int k = 0; k < n - 1; k ++ )
    {
    
    
        string next;
        for (int i = 0; i < cur.size();)
        {
    
    
            int j = i + 1;
            while (j < cur.size() && cur[j] == cur[i]) j ++ ;
            next += cur[i] + to_string(j - i);
            i = j;
        }
        cur = next;
    }
    cout << cur << endl;
    return 0;
}