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
