1140 Look-and-say Sequence (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

 

思路：用s保存上一次描述的字符串， 用s1保存这一次的字符串。 ch记录第一个不同字符， begin记录第一个不同字符的位置， j记录下一个不同字符开始的位置， 两者的差值就是相同字符的个数；

#include<iostream>
#include<string>
using namespace std;
int main(){
   int d, n;
   string s="";
   cin>>d>>n;
   char dig[]={'0','1','2','3','4','5','6','7','8','9'};
   s += dig[d];
   for(int i=1; i<n; i++){
       string s1="";
       char ch=s[0];
       int begin=0;
       for(int j=0; j<s.size();){
           while(s[j]==ch){j++;}
           int len=j-begin;
           begin=j;
           s1 += ch;
           s1 += dig[len];
           ch = s[j];
       }
       s = s1;
   }
   cout<<s<<endl;
  return 0;
}