pat1082. Read Number in Chinese (25)
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
数的读法转换,比较灵活。具体见代码注释。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
string dight[]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string id[]{"","Wan","Yi"};
string num[]{"","Shi","Bai","Qian"};
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
string s;
cin>>s;
reverse(s.begin(),s.end());
int i=s.length()-,j;
queue<string> q;
if(s[i]=='-'){
i--;
q.push("Fu");
}
if(s[i]==''){//特殊情况
q.push("ling");
i--;
}
for(;i>=;i--){
j=i;
if(i>=&&s[i]==''){
while(i>=&&s[i]==''){
i--;
}
i++;//保证i一定>=0
if(i/==j/){//0在同一区间
if(i%!=){//连续的0在区间中段
q.push("ling");
}
}
else{//多个0跨越区间
q.push(id[j/]);//只可能跨越万和个的区间
if(i!=){//最后一个0不是个位
q.push("ling");
}
}
}
else{
q.push(dight[s[i]-'']);
if(i%!=){//不是个位的情况
q.push(num[i%]);
}
}
if(i%==&&i>){//到了区间末尾
q.push(id[i/]);
}
}
cout<<q.front();
q.pop();
while(!q.empty()){
cout<<" "<<q.front();
q.pop();
}
cout<<endl;
return ;
}
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