1140 Look-and-say Sequence

Problem #

Source: [PAT 1140]

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

Solution #

• 题意 给你一个数字D和次数n，执行n次f操作，f操作是计算当前数字从前往后每个连续数字出现的次数，如1经行f操作为11，1121134进行f操作为1221123141
• 思路 以字符串的形式进行f操作

Code #

#include <iostream>
#include <algorithm>
#include <string>
#define maxsize 202
using namespace std;
int m, n;
string s, tmp;
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    cin >> s >> m;
    for (int i = 0; i < m-1; i++)
    {
        tmp = s[0];
        n = 1;
        int j;
        for (j = 1; j < s.size(); j++)
        {
            if (s[j] == s[j - 1])
                n++;
            else
            {
                tmp += ('0' + n);
                tmp += s[j];
                n = 1;
            }
        }
        if (j == s.size())
            tmp += ('0' + n);
        s = tmp;
    }
    cout << s << endl;
    return 0;
}