目录
题目
Input Specification:
Output Specification:
Sample Input:
Sample Output:
思路
代码
题目
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
思路
难度评级:⭐️
代码
#include <iostream>
#include <vector>using namespace std;int main(int argc, char** argv) {int d,n;cin>>d>>n;vector<int> pre,post;pre.push_back(d);for(int i=0;i<n-1;i++) {int x=pre[0],xn=0;// 遍历prefor(int j=0;j<pre.size();j++) {if(pre[j]==x) xn++;else {post.push_back(xn);x=pre[j];xn=1;}if(xn==1) post.push_back(x);} post.push_back(xn);pre=post;post.clear();}for(int i=0;i<pre.size();i++) cout<<pre[i];return 0;
}