pat1082. Read Number in Chinese (25)

1082. Read Number in Chinese (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

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 ;
 }