PAT甲级-1140. Look-and-say Sequence (20)（模拟）

港控/mmm° 2022-05-29 06:15 148阅读 0赞

# 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

题目意思：

对于样例 ：D, D1, D111, D113, D11231, D112213111,

D是首项“D”，其中D是一个给定的0~9的整数；

D1表明D中连续“D”有“1”个；

D111表明D1中连续“D”有“1”个，连续“1”有“1”个；

D113表明D111中连续“D”有“1”个，连续“1”有“3”个；

D11231表明D113中连续“D”有“1”个，连续“1”有“2”个，连续“3”有“1”个；

以此类推……

给定D，求第N项的表示方法。

解题思路：

用字符数组存储，依次遍历前一个字符，计数连续出现的字符，存入当前字符。

#include<bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define MAXN 1010
    string a[MAXN];
    
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        int d,n;
        cin>>d>>n;
        a[0]=char(d+'0');
        a[1]=a[0]+"1";
        for(int i=2; i<n; ++i)
        {
            string s=a[i-1];
            int cnt=1;
            int len=s.length();
            for(int j=1; j<len; ++j)//遍历前一个串
            {
    
                if(s[j]==s[j-1]) ++cnt;
                else
                {
                    a[i]+=s[j-1];//元素
                    a[i]+=char(cnt+'0');//个数
                    cnt=1;
                }
            }
            a[i]+=s[len-1];//加上最后一次的计数
            a[i]+=char(cnt+'0');
        }
        //for(int i=1; i<n; ++i)
        cout<<a[n-1]<<endl;
        return 0;
    }