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

  作者 CHEN, Yue 单位 浙江大学 代码长度限制 16 KB 时间限制 400 ms 内存限制

这道题是数数字，我们进行遍历字符串，数一共多少连续数字即可。

#include <iostream>
using namespace std;
string cal(string s) {
    string ans = "";
    char ch = s[0], cnt = 1;
    for(int i = 1; i < s.length(); i++) {
        if(s[i] != s[i - 1]) {
            ans += ch;
            ans += (cnt + '0');
            ch = s[i];
            cnt = 1;
        } else cnt++;
    }
    ans += ch;
    ans += (cnt + '0');
    return ans;
}
int main() {
    string D; int N;
    cin >> D >> N;
    while(--N) D = cal(D);
    cout << D;
    return 0;
}