PAT甲級1140 Look-and-say Sequence (20分) 模擬

1140 Look-and-say Sequence (20分)
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

這個題就是模擬吧
樣例的計算過程如下

最初是 1
然後有一個1 就是 11
然後有兩個1 就是12，然後一個1 一個2就是 1121
然後有兩個1 一個2 一個1 就是122111

我代碼可能麻煩，但是比較符合我思路，就是兩個變量，一個thisNum記錄當前計數的是哪個 ，比如我們是計算有幾個1 還是幾個2 ，這裏的1 和2 就是thisNum要保存的數字。 然後一個number變量，記錄thisNum當前的個數

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;

int main()
{
    int D,N;
    cin>>D>>N;
    vector<int> vec;
    vec.push_back(D);

    for(int i=0; i<N-1; i++)
    {
        vector<int> out;
        int thisNum=-1;
        int number=0;
        for(int q=0; q<vec.size(); q++)
        {
            if(thisNum==-1)
            {
                thisNum=vec[q];
                number=1;
            }
            else
            {
                if(vec[q]==thisNum)
                {
                    number++;
                }
                else
                {
                    out.push_back(thisNum);
                    out.push_back(number);
                   thisNum=vec[q];
                number=1;
                }
            }
        }
        if(thisNum!=-1)
        {
            out.push_back(thisNum);
            out.push_back(number);
            thisNum=-1;
        }
        vec=out;
        for(int i=0; i<out.size(); i++)
        {
            cout<<out[i];
        }
        cout<<endl;
    }
    return 0;
}