1140 Look-and-say Sequence （20 分）简单模拟

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
思路
　　为了求第n+1个，要对第n个进行遍历。对于每一个遍历到的数字，首先把该数字保存，然后计数该数字连续相同的个数。
把个数也保存当这个遍历结束的时候，第n+1个就求出来了。

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


int main()
{
    int d,n;
    cin>>d>>n;
    vector<int> temp;
    temp.push_back(d);
    vector<int>print;
    for(int i=0;i<n-1;i++)
    {
        for(int j=0;j<temp.size();)
        {
            int m=j,coun=0;
            print.push_back(temp[j]);
            while(m<temp.size())
            {
                if(temp[m]==temp[j])
                {
                    m++;
                    coun++;
                }
                else
                    break;
            }
            j=m;
            print.push_back(coun);
        }
        temp=print;
        print.clear();
    }
    print=temp;
    for(auto it:print)
        cout<<it;
    return 0;
}