1140 Look-and-say Sequence

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

 

题意：

　　用后一个数描述前一个数，例如：1，11, 12, 1121，第一个数是给定的，第二个数是说：第一个数由1个1组成；第三个数是说：第二个数由2个1组成；第四个数是说：第三个数由1个1和1个2组成。

思路：

　　首先将数字转成字符串，末尾加一个空格（方便输出），ans用来存储这次生成的字符串。注意有组测试是测的第一个数字，直接输出就行了。

 

Code：

#include <iostream>
#include <string>

using namespace std;

int main() {
    int D, N, count;
    cin >> D >> N;
    if (N == 1) {
        cout << D << endl;
        return 0;
    }
    string ans = to_string(D) + "1";
    string str = ans + " ";
    for (int i = 3; i <= N; ++i) {
        count = 1;
        ans = "";
        for (int j = 1; j < str.length(); ++j) {
            if (str[j] == str[j - 1])
                count++;
            else {
                ans += str[j - 1] + to_string(count);
                count = 1;
            }
        }
        str = ans + " ";
    }
    cout << ans << endl;
    return 0;
}