#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(zhì)就是按照順序數(shù)出每個(gè)數(shù)的出現(xiàn)次數(shù)，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(jié)規(guī)律：將字符串分成多個(gè)部分，每一部分的值相同，數(shù)出每一部分對應(yīng)元素個(gè)數(shù)num,組成對應(yīng)值val新的子串: num val例如有11個(gè)2，那么變成112132221:11133211輸入:123456數(shù)出:111211211111221312211關(guān)鍵:1總結(jié)規(guī)律：將字符串分成多個(gè)部分，每一部分的值相同，數(shù)出每一部分對應(yīng)元素個(gè)數(shù)num,組成對應(yīng)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(jì)數(shù)器變成1，當(dāng)前數(shù)出現(xiàn)1次			}		}		//最后一個(gè)元素還沒有統(tǒng)計(jì)，因?yàn)橹爸皇谴娣派弦淮蔚慕Y(jié)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(jìn)行數(shù)		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(jì)數(shù)器變成1，當(dāng)前數(shù)出現(xiàn)1次			}		}		//最后一個(gè)元素還沒有統(tǒng)計(jì)，因?yàn)橹爸皇谴娣派弦淮蔚慕Y(jié)果		Result result(count , value);		results.push_back(result);		//拼接結(jié)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(gè)，初始元素為1    string countAndSay(int n) {		if(n <= 0)		{			return "";		}        //需要將整數(shù)轉(zhuǎn)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;}