PAT_A1140#Look-and-say Sequence-白红宇

PAT_A1140#Look-and-say Sequence

阅读量：5860 次

发布时间：2019-06-19

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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 ( $\leq 40), separated by a space.$

Output Specification:

Print in a line the Nth number in a look-and-say sequence of D.

Sample Input:

1 8

Sample Output:

1123123111

Keys:

简单模拟

Attention:

这种小题有时候还挺头疼的-，-

Code:

1 /* 2 Data: 2019-05-24 10:44:34 3 Problem: PAT_A1140#Look-and-say Sequence 4 AC: 44:30 5  6 题目大意： 7 观察并说出响应的序列； 8 比如给出第一个数字D，第二个数字为D1（D有1个） 9 第三个数字为D111（D有1个，1有1个）；10 第四个数字为D113（D有1个，1有3个）；11 以此类推....12 输入：13 初始数字D，和轮次N14 输出：15 第N轮响应的序列16 */17 18 #include
     
      19 #include
      
       20 #include
       
        21 using namespace std;22 23 int main()24 {25 #ifdef    ONLINE_JUDGE26 #else27     freopen("Test.txt", "r", stdin);28 #endif29 30     int n;31     string s;32     cin >> s >> n;33     for(int i=1; i