## 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;
}
```