PAT Advanced 1140 Look-and-say Sequence (20 分)

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

 

乙级真题

 

#include <iostream>
using namespace std;
string coun(string str){
    string res="";int coun=1;
    for(int i=1;i<str.length();i++){
        if(str[i]==str[i-1]) coun++;
        else {
            res+=(str[i-1]);
            res+=(coun+'0');
            coun=1;
        }
    }
    res+=str[str.length()-1];res+=(coun+'0');
    return res;
}
int main()
{
    string A;int B;
    cin>>A>>B;
    for(int i=1;i<B;i++) A=coun(A);
    cout<<A;
    system("pause");
    return 0;
}

 

posted @ 2019-11-06 16:42  SteveYu  阅读(147)  评论(0编辑  收藏  举报