#include <iostream>#include <stdio.h>#include <vector>#include <sstream>using namespace std;/*問題:The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 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, generate the nth sequence.Note: The sequence of integers will be rePResented as a string.分析：此題本質就是按照順序數出每個數的出現次數，然后翻譯層字符串。1211：One 1,  One 2, Two 1,得到 111221111221:Three 1,Two 2,One 1,       31,22,11得到: 312211312211:One 3,Two 2,Two 1       13,22,21得到132221總結規律：將字符串分成多個部分，每一部分的值相同，數出每一部分對應元素個數num,組成對應值val新的子串: num val例如有11個2，那么變成112132221:11133211輸入:123456數出:111211211111221312211關鍵:1總結規律：將字符串分成多個部分，每一部分的值相同，數出每一部分對應元素個數num,組成對應值val新的子串: num val2 		for(int i = 1 ; i < len ; i++)		{			//找到相同元素			if(str.at(i) == value)			{				count++;			}			else			{				Result result(count , value);				results.push_back(result);				value = str.at(i);				count = 1;//計數器變成1，當前數出現1次			}		}		//最后一個元素還沒有統計，因為之前只是存放上一次的結果		Result result(count , value);		results.push_back(result);*/typedef struct Result{	Result(int times , char value):_times(times), _value(value){}	int _times;	char _value;}Result;class Solution {public:	string getNum(string str)	{		if(str.empty())		{			return "";		}		int len = str.length();		//開始進行數		char value = str.at(0);		int count = 1;		vector<Result> results;		for(int i = 1 ; i < len ; i++)		{			//找到相同元素			if(str.at(i) == value)			{				count++;			}			else			{				Result result(count , value);				results.push_back(result);				value = str.at(i);				count = 1;//計數器變成1，當前數出現1次			}		}		//最后一個元素還沒有統計，因為之前只是存放上一次的結果		Result result(count , value);		results.push_back(result);		//拼接結果		int size = results.size();		stringstream resultStream;		for(int i = 0 ; i < size ; i++)		{			resultStream << results.at(i)._times << results.at(i)._value;		}		string realResult = resultStream.str();		return realResult;	}	//注意這里的n為第幾個，初始元素為1    string countAndSay(int n) {		if(n <= 0)		{			return "";		}        //需要將整數轉換成字符串		string num("1");		if(1 == n)		{			return num;		}		string result;		for(int i = 2 ; i <= n ; i++)		{			result = getNum(num);			num = result;		}		return result;    }};void process(){	int n;	while(cin >> n)	{		Solution solution;		string result = solution.countAndSay(n);		cout << result << endl;	}}int main(int argc , char* argv[]){	process();	getchar();	return 0;}