1082 Read Number in Chinese (25 分)
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
题目大意:将数字转换成中文读法。注意stoi的使用,以及substr(i,j)的结果是得到 i 到 j-1 的子字符串。
自己需要手动输入一些测试用例,如:
测试用例: 1010010 yi Bai ling yi Wan ling yi Shi
1000010 yi Bai Wan ling yi Shi
100000000 yi Yi
0000 ling
#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int main()
{
string str;
cin>>str;
string a[]={"","yi","er","san","si","wu","liu","qi","ba","jiu"};
string b[] = { "", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi" };
vector<string> vec;
for(int i=str.size()-1,k=0;i>=0;i--,k++)
{
if(str[i] == '0'){
if(k == 4){
string s = str.substr(0,i+1);//判断是否需要加"Wan",比如一亿
int n = abs(stoi(s));
if(n % 10000){//如101000000,需要加,100000010不需要
vec.push_back("Wan");
}
}
if(vec.empty() || vec[vec.size()-1] == "ling" || vec[vec.size()-1] == "Wan")
{
continue;
}
vec.push_back("ling");
continue;
}
if(str[i] == '-'){
vec.push_back("Fu");
break;
}
if(k)
vec.push_back(b[k]);
vec.push_back(a[str[i]-'0']);
}
for(int i=vec.size()-1;i>=0;i--)
{
cout<<vec[i];
if(i != 0)
cout<<" ";
}
if(vec.size() == 0)
cout<<"ling";
return 0;
}