PAT 1140 Look-and-say Sequence [比较]

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

 题目大意：描述序列，每一个序列都是描述前一个序列的。要求给出第n个序列。

//本来一看差点懵了，但是告诉自己这个我肯定会做，然后就写出来啦，就是简单地找规律而已。

#include <iostream>
#include <algorithm>
#include <vector>
#include<string.h>
#include<string>
#include<cstdio>
using namespace std;

string get(int c){//将计数转换为字符串
    string s;
    while(c!=0){
        s+=(c%10+'0');
        c/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}
int main()
{
    string s1,s2;
    int n;
    cin>>s1>>n;
    //s1=s1+"1";
    for(int i=1;i<n;i++){
        int ct=1;
        for(int j=0;j<s1.size();j++){
            while(s1[j]==s1[j+1]&&j<s1.size()-1){
                ct++;j++;
            }
            s2+=s1[j]+get(ct);
            ct=1;//这里ct要转化为字符串。
        }
        s1=s2;
        s2="";
    }
    cout<<s1;

    return 0;
}

1.一开始直接用ct+'0'出现了乱码的情况，然后就简单地写了一个函数，转换为字符串，况且对于ct>10的那种也没法通过+‘0’直接转换了

2.后来提交有一个测试点过不去，后来思考发现是因为自己一开始一进来就把s1+“1”，这样是不对的。修改了一下就可以了。