藍橋題目：B-20、數的讀法

基本思想：從最小位，即字符串的末端開始計算，每四個進行統計，插入位數和讀法。

#include <iostream>
#include <cstring>

using namespace std;

int main() {
	string n;
	cin >> n;
	
	char *read[] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
	char *sign[] = {"","shi","bai","qian","wan","yi"};//開頭多加一個“”使座標更方便 
	string rst;
	int l = n.length();
	int flag = 0;//用來標記這是四位數中的第幾個 
	for(int i = l-1; i >= 0; --i){
		
		if(flag % 4 == 0){//若是第一位 
			
			if(i == l-1){//判斷是否是整體第一個數 
				flag++;
			}else{
				if( ( (l-i) / 4 ) % 2){
					if(n[i] == '0' && n[i-1] == '0' && n[i-2] == '0' && n[i-3] == '0'){//四位數全爲0直接跳過 
						i -= 3;
						continue;
					}
					rst = sign[4] + rst;//前加萬
					rst = " " + rst;
				}else{
					rst = sign[5] + rst;//前加億 
					rst = " " + rst;
				}
				flag++;
			}
			if(n[i] == '0'){
				continue;
			}
			
		}else{
			if(n[i] == '0'){
				if(n[i+1] != '0'){//四位數中間非連續的0需讀出 
					rst = read[0] + rst;
					rst = " " + rst;
				}
				flag++;
				continue;
			}
			rst = sign[flag % 4] + rst;//前加十百千 
			rst = " " + rst;
			flag++;
		}
		
		rst = read[ n[i] - '0' ] + rst;
		rst = " " + rst;
		
	}
	
	rst.erase(0,1);
	int index = 0;
	if(rst.find("yi shi",index) == 0)
		rst.erase(0,3); 
	cout << rst; 
	
	return 0;
}