题目：

The count-and-say sequence is the sequence of integers with the first five terms as following:

  1. 1
  2. 11
  3. 21
  4. 1211
  5. 111221

1 is read off as “one 1” or 11.
11 is read off as “two 1s” or 21.
21 is read off as “one 2, then one 1” or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

题意：

字符串读法问题，第二个字符串是第一个字符串的读法，第三个字符串是第二个字符串的读法。
规则大致就是说 几个几。
比如1211，他是1个1、1个2、2个1，故读作111221.

思路：

第一个字符串为1，共变化n-1次。每次从左向右扫描，与前一字符相同则计数+1，不同则将计数值和字符串加在字符串上。

代码：

string countAndSay(int n) {
	string res = "1";
	while (--n) {
		string temp;
		int count = 1;
		char now = res[0];
		for (auto it = res.begin() + 1; it != res.end(); ++it) {
			if (*it == now)
				++count;
			else {
				temp += ('0' + count);
				temp += now;
				now = *it;
				count = 1;
			}		
		}
		res = temp;
		res += ('0' + count);
		res += now;
	}
	return res;
}