1140 Look-and-say Sequence

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

题意

给定一个 D 和 n ，按照 D 需要输出第 n 项外观序列，而外观序列的规则如下：

第 i 项序列要根据其前一项的序列生成，例如 D 等于 1 ，则第 2 项为 11 ，意思是前一项中 1 出现了 1 次；第 3 项为 12 ，即其前一项中 1 出现了 2 次；第 3 项为 1121 ，即其前一项中 1 出现了 1 次，2 出现了 1 次；第 4 项为 122111 ，即其前一项 1 先连续出现了 2 次，接着2 出现了 1 次，然后 1 又出现了 1 次。

所以可以发现，每项序列是根据其上一个序列的值中每个元素连续的个数来生成的。

思路

具体思路如下：

1.输入起始序列 num 以及需要输出的项数 n 。

2.模拟题目意思，用双指针来生成每项序列，p 指向当前元素，而 q 指向与 p 指向元素相同的连续的最后一个元素的下一个位置，这样 q-p 就是 p 指向元素连续的个数。

3.输出第计算出来的第 n 项序列。

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string num;
    int n;
    cin >> num >> n;
    //进行n-1次转换
    for (int i = 1; i < n; i++)
    {
        int len = num.size();
        int p = 0;
        string str = "";
        while (p < len)
        {
            int q = p + 1;
            //找出连续相同的字段
            while (q < len && num[q] == num[p])    q++;
            str += num[p] + to_string(q - p);
            p = q;
        }
        num = str;
    }
    //输出第n项序列
    cout << num << endl;
    return 0;
}