PAT A1140 Look-and-say Sequence （20 分）——数学题

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

 #include <stdio.h>
 #include <algorithm>
 #include <iostream>
 #include <map>
 #include <vector>
 #include <set>
 using namespace std;
 const int maxn=;
 int n,m,k;
 int seq[][maxn];
 int cnt;
 int main(){
     scanf("%d %d",&n,&m);
     seq[][]=n;
     if(m==)printf("%d",n);
     cnt=;
     for(int i=;i<m;i++){
         k=;
         int now=seq[i-][],num=;
         for(int j=;j<cnt;j++){
             if(seq[i-][j]==now){
                 num++;
             }
             else{
                 seq[i][k]=now;
                 seq[i][k+]=num;
                 now=seq[i-][j];
                 num=;
                 k+=;
             }
         }
         seq[i][k]=now;
         seq[i][k+]=num;
         k+=;
         cnt=k;
     }
     for(int j=;j<k;j++){
         printf("%d",seq[m-][j]);
     }
 }

注意点：题目看了半天没懂，后来看懂了点理解的是前一个序列有几个什么然后输出，然后第六个样例怎么看都不对。看了大佬的思路，原来是有几个连续的数字，然后输出来。那既然n最大就40，设个二维数组直接枚举就好了，maxn一开始只设了10010，发现最后一个测试点错了，最后一个测试点应该是n=40，结果很大，maxn为1e5就够了