1082 Read Number in Chinese (25)
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
作者: CHEN, Yue
单位: PAT联盟
时间限制: 400ms
内存限制: 64MB
代码长度限制: 16KB
好菜啊,没想出来这个题怎么写,卡在不知道怎么简洁的输出“Yi”,“Wan”这些表示数量的单词上
看了一下题解,因为最多只有9位数,所以可以把整个数字分为三节(如果有的话):个位段(1~4位),万位段(5~8位),亿位段(第9位),针对每一段进行处理,从最高位开始。
我们设两个指针left和right,left指向字符串首位,right指向末位,按段处理数据
AC代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
string num[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string numstr[5]={"Shi","Bai","Qian","Wan","Yi"};
string str;
cin>>str;
int len=str.length();
int left=0,right=len-1;//left和right分别指向字符串首尾元素
if(str[0]=='-')
{
cout<<"Fu";
left++;
}
while(left+4<=right)
right-=4;
//cout<<"left= "<<left<<endl;
//cout<<"right= "<<right<<endl;
while(left<len)
{
bool flag=false;
bool isprint=false;
while(left<=right)
{
if(left>0&&str[left]=='0')//如果有连续的0就会一直执行
flag=true;
else//遇到不为0的数
{
if(flag==true)//存在累积的0
{
cout<<" ling";
flag=false;
}
//非首位后面每一位要多输出一个空格
if(left>0)
cout<<" ";
cout<<num[str[left]-'0'];
isprint=true;
if(left!=right)//输出十百千位
cout<<" "<<numstr[right-left-1];//仔细体会right-left-1
}
left++;//left右移一位
}
if(isprint==true&&right!=len-1)//不是个位就输出万或亿
cout<<" "<<numstr[(len-1-right)/4+2];
//right距离末尾的距离只有0,4,8三种可能
//right!=len-1排除了1~5位数的情况(距离为0)
right+=4;//右移4位,输出下一节
}
return 0;
}
//本题最多9位数,也就是最大的数是x亿
//我们可以把整个数字划分为个节,万节,亿节
//我们从最高位的亿节开始输出,依次到万节,个节