PAT 甲级 1140 Look-and-say Sequence (20分)

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

题目大意：第n+1个串是第n个串的描述， 若第n个串是11122234，那么第n+1个串就是13233141，即1有3个、2有3个、3有1个、4有1个。

样例解释：暂无

【AC代码】：

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n; string s; cin >> s >> n;
    for(int i = 2; i <= n; i++)
    {
        string k; int top = 1;
        for(int j = 1; j <= s.length(); j++)
        {
            if(s[j] == s[j-1]) top++;
            else{
                k += s[j-1] + to_string(top);
                top = 1;
            }
        }
        s = k;
    }
    cout << s;
    return 0;
}