PAT 1140 Look-and-say Sequence

1140 Look-and-say Sequence

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

总结：这道题目之前在乙级已经做过了，所以没有多大的难度，不过有一个点需要注意（写在代码里了，自己猜测的原因）

代码：

#include 
using namespace std;
 
int main(){
    string s;
    int n;
    cin >> s >> n;
    
    while(n-->1){
        string t;
        for(int i=0;isize();){
            int cot=0,j=0;
            for(j=i;jsize() && s[i]==s[j];j++)   cot++;
            t+=s[i]+to_string(cot);
            //t=t+s[i]+char(cot+'0');这样耗时会长很多，因为这里重新定义了一个临时变量t
            i+=cot;
        }
        s=t;
    }
    
    cout << s << endl;
    return 0;
}

 好好学习，天天向上！

我要考研！