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=100100;
int n,m,k;
int seq[50][maxn];
int cnt;
int main(){
    scanf("%d %d",&n,&m);
    seq[0][0]=n;
    if(m==1)printf("%d",n);
    cnt=1;
    for(int i=1;i<m;i++){
        k=0;
        int now=seq[i-1][0],num=0;
        for(int j=0;j<cnt;j++){
            if(seq[i-1][j]==now){
                num++;
            }
            else{
                seq[i][k]=now;
                seq[i][k+1]=num;
                now=seq[i-1][j];
                num=1;
                k+=2;
            }
        }
        seq[i][k]=now;
        seq[i][k+1]=num;
        k+=2;
        cnt=k;
    }
    for(int j=0;j<k;j++){
        printf("%d",seq[m-1][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