PAT_A1140#Look-and-say Sequence

Source:

PAT A1140 Look-and-say Sequence (20 分)

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 (≤ 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-，-
• to_string()

Code:

 /*
 Data: 2019-05-24 10:44:34
 Problem: PAT_A1140#Look-and-say Sequence
 AC: 17:49

 题目大意：
 观察并说出相应的序列；
 比如给出第一个数字D，第二个数字为D1（D有1个）
 第三个数字为D111（D有1个，1有1个）；
 第四个数字为D113（D有1个，1有3个）；
 以此类推....
 输入：
 初始数字D，和轮次N
 输出：
 第N轮相应的序列
 */
 #include<cstdio>
 #include<string>
 #include<iostream>
 using namespace std;

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n;
     string s;
     cin >> s >> n;
     for(int k=; k<n; k++)
     {
         string t="";
         for(int i=; i<s.size(); i++)
         {
             int cnt=;
             while(i+<s.size() && s[i]==s[i+])
             {
                 cnt++;
                 i++;
             }
             t += (s.substr(i,)+to_string(cnt));
         }
         s=t;
     }
     cout << s;

     return ;
 }