1082. Read Number in Chinese (25)

最新推荐文章于 2021-09-04 23:23:15 发布

原创最新推荐文章于 2021-09-04 23:23:15 发布 · 470 阅读

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1.需要分类考虑的情况比较多

2.123456789可以分为 1 2345 6789 ，其中2345和6789的处理方法相同，不同点就是2345处理完后需要加上Wan

3.需要处理0等特殊情况

4下面有些特别的测试例子，可以参考：

/*
-123456789
123400000
123000000
120000000
100000000
123000010
-23456789
23400000
23000000
20000000
00000000
23000010
*/

//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;
/*
-123456789
 123400000
 123000000
 120000000
 100000000
 123000010
-23456789
 23400000
 23000000
 20000000
 00000000
 23000010
*/
string num2Chinese[] = { "ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };
string deleteZero(string str)
{
	string tmp = "";
	bool firstZero = true;
	for (int i = 0; i < str.size(); i++)
	{
		if (str[i] == '0'&&firstZero)
			;
		else
		{
			firstZero = false;
			tmp += str[i];
		}
	}
	return tmp;
}
string digit1(string str)
{//只有一位数字的时候
	if (str == "") return str;
	else return num2Chinese[str[0] - '0'];
}
string digit2(string str)
{//两位数组，10为yi Shi，11为yi Shi yi，处理这两种特殊情况：1）个位为0：2）各位不为0
	string strTmp = num2Chinese[str[0] - '0'] + " Shi";
	string next = deleteZero(str.substr(1));
	if (next.size() == 1)
		return strTmp + " " + digit1(next);
	else return strTmp;
}
string digit3(string str)
{//100,101,110,111处理三种特殊情况，1）十位不为0；2）十位为0，个位不为0；3）十位、个位均为0
	string strTmp = num2Chinese[str[0] - '0'] + " Bai";
	string next = deleteZero(str.substr(1));
	if (next.size() == 2)
		return strTmp + " " + digit2(next);
	else if (next.size() == 1)
		return strTmp + " ling " + digit1(next);
	else return strTmp;
}
string digit4(string str)
{//1000，1001，1011，1111处理四种特殊情况，1）百位不为0；2）百位为0，十位不为0，个位不为0；3）百位、十位均为0，个位不为0；4）百位、十位、个位均为0
	string strTmp = num2Chinese[str[0] - '0'] + " Qian";
	string next = deleteZero(str.substr(1));
	if (next.size() == 3)
		return strTmp + " " + digit3(next);
	else if (next.size() == 2)
		return strTmp + " ling" + digit2(next);
	else if (next.size() == 1)
		return strTmp + " ling" + digit1(next);
	else return strTmp;
}
string digitProc(string str)
{
	switch (str.size())
	{
	case 0:return digit1(str);
	case 1:return digit1(str);
	case 2:return digit2(str);
	case 3:return digit3(str);
	case 4:return digit4(str);
	default:
		return "";
	}
}

int main(void)
{
	string str;
	cin >> str;
	bool sign = true;
	if (str[0] == '-')
	{
		sign = false;
		str = str.substr(1);
	}
	str = deleteZero(str);
	string ans = "";
	if (str.size() == 9)
	{//包括亿位
		string a = str.substr(0, 1);
		string b = str.substr(1, 4);
		string c = str.substr(5, 4);
		ans = num2Chinese[a[0] - '0'] + " Yi";
		string tmpB = deleteZero(b);
		string tmpC = deleteZero(c);
		if (tmpB.size() == 4 && tmpC.size() == 4)
			ans += " " + digitProc(tmpB) + " Wan " + digitProc(tmpC);
		else if (tmpB.size() == 4 && tmpC.size() == 0)
			ans += " " + digitProc(tmpB) + " Wan ";
		else if (tmpB.size() == 4 && tmpC.size() < 4)
			ans += " " + digitProc(tmpB) + " Wan ling " + digitProc(tmpC);
		else if (tmpB.size() == 0 && tmpC.size() == 0)
			;
		else if (tmpB.size() == 0 && tmpC.size() <= 4)
			ans += " ling " + digitProc(tmpC);
		else if (tmpB.size() < 4 && tmpC.size() == 0)
			ans += " ling" + digitProc(tmpB) + " Wan ";
	}
	else if (str.size() < 9 && str.size() > 4)
	{//不包括亿位,包括万位
		string b = str.substr(0, str.size() - 4);
		string c = str.substr(str.size() - 4);
		string tmpB = deleteZero(b);
		string tmpC = deleteZero(c);
		if (tmpB.size() == b.size() && tmpC.size() == c.size())
			ans += digitProc(tmpB) + " Wan " + digitProc(tmpC);
		else if (tmpB.size() == 0 && tmpC.size() == 0)
			;
		else if (tmpB.size() == 0 && tmpC.size() <= 4)
			ans += digitProc(tmpC);
		else if (tmpB.size() == 4 && tmpC.size() == 0)
			ans += digitProc(tmpB) + " Wan ";
		else if (tmpB.size() < 4 && tmpC.size() == 0)
			ans += digitProc(tmpB) + " Wan ";
		else if (tmpB.size() < 4 && tmpC.size() < 4)
			ans += digitProc(tmpB) + " Wan ling " + digitProc(tmpC);
		else if (tmpB.size() == 4 && tmpC.size() < 4)
			ans += digitProc(tmpB) + " Wan ling " + digitProc(tmpC);
	}
	else if (str.size() <= 4 && str.size() != 0)
	{
		ans = digitProc(str);
	}
	else if (str.size() == 0)
		ans = "ling";
	if (!sign)
		ans = "Fu " + ans;
	cout << ans << endl;

	return 0;
}