[PAT] C++ 1140. Look-and-say Sequence (20)

本文链接：https://blog.youkuaiyun.com/lv1224/article/details/79605291

本文介绍了一种名为“看说”数列的生成算法，该数列通过描述前一个数列来生成下一个数列。文章提供了一个具体的C++实现案例，输入为一个0到9之间的数字D和一个正整数N，输出为该数列的第N项。

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题目

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

代码

#include <bits/stdc++.h>
using namespace std;

int main(){
    string s,res;
    int n,m;
    scanf("%d%d",&n,&m);
    s=n+'0';
    for(int i=1;i<m;i++){
        res="";
        int j=0;
        while(j<s.size()){
            int k=j+1;
            while(k<s.size()&&s[k]==s[j])
                k++;    
            res+=s[j];
            res+=k-j+'0';
            j=k;
        }
        s=res;
    }
    if(res.size()==0) cout<<s<<endl;
    else cout<<res<<endl;
    return 0;
}