PAT甲级——A1140 LookAndSaySequence【20】

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

Solution：
　　能不能读懂这道题是个关键
　　举例序列：
　　　　D　　D1　　D111　　D113　　D11231
　　解释：
　　　　D 怎么得到 D1
　　　　　　D中有 D 1个  ==》  D1
　　　　D1怎么得到D111  
　　　　　　D1中有D 1个，1 1个  ==> D1 11    D111
　　　　D111怎么得到D113
　　　　　　D111中有D1个，1 3个  ==》 D1 13    D113
记住，D是输入的那个数字！！！！！我就没想通这点，坑死了

 #include <iostream>
 #include <string>
 using namespace std;
 int main()
 {
     string s;
     int n;
     cin >> s >> n;
     for (int cnt = ; cnt < n; ++cnt)
     {
         string str = "";
         int k;
         for (int i = ; i < s.length(); i=k)//计算各个字母出现的个数, i=k,从不重复的字母开始
         {
             for (k = i; k < s.length() && s[k] == s[i]; ++k);//计算相同的字母个数
             str += s[i] + to_string(k - i);//将计算的该字母和其相同的次数算入
         }
         s = str;
     }
     cout << s;
     return ;
 }