A1082 Read Number in Chinese (25 point(s))

数字转字符

1. 原文

原题

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.

Sample Input 1:

-123456789

Sample output 1:

Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu

Sample Input 2:

100800

Sample output 2:

yi Shi Wan ling ba Bai

2. 解析

• 数字转数组

  do{
      	num[idx++]=n%10;
      	n/=10;
  }while(n!=0);

• 数组转字符串

  if (i!=0&&(num[i]!=0||i==4||i==8)){
      ans[t++]=shi[i];
  }
  ans[t++]=ge[num[i]];

3. AC代码

#include<iostream>
#include<string>
using namespace std;
string ge[]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string shi[]={"","Shi","Bai","Qian","Wan","Shi","Bai","Qian","Yi"};
int num[20]={};
string ans[20];
int main(){
	int n;
	cin>>n;
	if (n<0)
	{
		cout<<"Fu ";
		n=-n;
	}else if(n==0){
    cout<<"ling"<<endl;
  }
	int idx = 0;
	do{
		num[idx++]=n%10;
		n/=10;
	}while(n!=0);
	int i = 0;
	for (; i < idx; ++i)
	{
		if (num[i]!=0)
		{
			break;
		}
	}
	int index = 0;
	for (; i < idx; ++i)
	{
		if (i!=0&&(num[i]!=0||i==4||i==8))
		{
			ans[index++]=shi[i];
		}
		ans[index++]=ge[num[i]];
	}
	i = index - 1;
	for (; i >= 0; i--)
	{
		int cnt = 0;
		while(ans[i]=="ling"&&i!=0){
			cnt++;
			i--;
		}
    //万节所有位位0，不能输出wan
		if (cnt>0&&ans[i]!="Wan")
		{
			cout<<"ling ";
		}
		cout<<ans[i];
		if (i>0)
		{
			cout<<" ";
		}else{
			cout<<endl;
		}	
	}


	
	return 0;
}