PAT_A1140#Look-and-say Sequence

Source:

PAT A1140 Look-and-say Sequence (20 分)

Description:

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

Keys:

• 简单模拟

Attention:

• 这种小题有时候还挺头疼的-，-
• to_string()

Code:

/*
Data: 2019-05-24 10:44:34
Problem: PAT_A1140#Look-and-say Sequence
AC: 17:49

题目大意：
观察并说出相应的序列；
比如给出第一个数字D，第二个数字为D1（D有1个）
第三个数字为D111（D有1个，1有1个）；
第四个数字为D113（D有1个，1有3个）；
以此类推....
输入：
初始数字D，和轮次N
输出：
第N轮相应的序列
*/
#include<cstdio>
#include<string>
#include<iostream>
using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE

    int n;
    string s;
    cin >> s >> n;
    for(int k=1; k<n; k++)
    {
        string t="";
        for(int i=0; i<s.size(); i++)
        {
            int cnt=1;
            while(i+1<s.size() && s[i]==s[i+1])
            {
                cnt++;
                i++;
            }
            t += (s.substr(i,1)+to_string(cnt));
        }
        s=t;
    }
    cout << s;

    return 0;
}