Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way. Output Fu first if it is negative. For example, -123456789 is read as Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu. Note: zero (ling) must be handled correctly according to the Chinese tradition. For example, 100800 is yi Shi Wan ling ba Bai

Input Specification:

Each input file contains one test case, which gives an integer with no more than 9 digits.

Output Specification:

For each test case, print in a line the Chinese way of reading the number. The characters are separated by a space and there must be no extra space at the end of the line.

解答：這道題我剛開始的解法是：對資料從低位開始，每四位進行劃分，然後分別求解，過程相當複雜，而且有一個用例過不了。

後來我借鑑了一位博主的思路，先對特殊情況進行判定，然後用棧從低位開始處理，每位的權重用陣列表示，最終得到結果，這是我遇到的最好的答案了。

不過他對特殊情況的判定只有0，然而如果是100000000，會將最後的Wan輸入，雖然測試用例過了，但補充下演算法更完整。

AC程式碼如下：

#include<iostream>
#include<string>
#include<stack>
#include<cstdlib>

using namespace std;

char digit[10][5] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
char weight[10][5] = {"", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi"};

int main()
{
	string n;
	cin >> n;
	//特殊情況0和100000000 
	if(n.size() == 1 && n[0] - '0' == 0){
		printf("%s\n", digit[0]);
		return 0;
	}
	if(n.size() == 9 && atoi(n.c_str()) == 100000000){
		printf("yi Yi\n");
		return 0;
	}
	
	if(n[0] == '-'){
		printf("Fu ");
		n.erase(0, 1);
	}
	
	stack<string> sta;
	int zeroFlag = 0;
	for(int i = n.size() - 1, j = 0; i >= 0; --i, j++){
		if(n[i] == '0'){
			if( zeroFlag == 1 && n[i+1] != '0' ){
				sta.push(digit[0]);
				zeroFlag = 0;
			}
			if( j == 4 ){
				sta.push(weight[j]);
			}
		}
		else{
			if(j != 0)  //只有j != 0時才將權重加入 
				sta.push(weight[j]);
			sta.push(digit[n[i] - '0']);
			zeroFlag = 1;
		}
	}
	//輸出結果
	int first = 1;
	while(!sta.empty()){
		if(first == 1){
			cout << sta.top();
			first = 0;
		}
		else{
			cout << " " << sta.top();
		}
		sta.pop(); 
	} 
	return 0;
}