PAT甲级：1136 A Delayed Palindrome (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

一道经典的滑动窗口题目。而我遇到了一个坑点。

• ans = ans + d[q] + to_string(p - q);
• ans += d[q] + to_string(p - q);

咋一看这两句是同一个意思，但是第一句在PAT测试点3是超时的。

教训就是，别闲着没事乱写，老老实实写+= ，明明就更简单是不是？

具体原因不太清楚，反正遇上了，就试试换成+= 看对不对？

#include <iostream>
#include <string>
using namespace std;
int main(){
    int n = 0;
    string d;
    d.resize(1);
    scanf("%s%d", &d[0], &n);
    for(int i = 1; i < n; i++){
        string ans;
        int p = 1, q = 0;
        d = d + "@";
        while(p < d.size()){
            if(d[p] != d[q]){
                //ans = ans + d[q] + to_string(p - q);
                ans += d[q] + to_string(p - q);
                q = p;
            }
            p++;
        }
        d = ans;
    }
    printf("%s", d.c_str());
    return 0;
}