1140. Look-and-say Sequence (20)

最新推荐文章于 2019-11-01 07:43:19 发布

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最新推荐文章于 2019-11-01 07:43:19 发布

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分类专栏： PAT_Advance 文章标签： PAT Advance 字符串

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本文介绍了一种基于描述的数列生成算法——看说数列。通过算法实现，可以生成给定数字D的第N个看说数列。文章包含完整的C++代码示例，演示了如何从初始数字开始，逐步生成后续数列。

1140. Look-and-say Sequence (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题目大意：第n+1个串是第n个串的描述。比如第n个串是11122234（随便写的），那么第n+1个串就是13233141，这个意思就是在第n个串中 1有3个， 2有3个， 3有1个， 4有1个，所以是13233141。

#include<bits/stdc++.h>
using namespace std;
char d;
int n,len;
string s,temp,ans;
int main(){
    cin>>d>>n;
    s=d;
    int cnt=1,num=0;
    char t;
    while(cnt<n){
        s+="a";
        cnt++;
        len=s.length();
        num=0;
        temp="";
        t=s[0];
        for(int i=0;i<len;i++){
           if(t==s[i])    num++;
            else {
                temp+=t;
                temp+=(char)(num+'0');
                num=0;
                t=s[i];
                i--;
            }
        }
        s=temp;
    }
    cout<<s<<endl;
    return 0;
}