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[10]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
string id[3]{"","Wan","Yi"};
string num[4]{"","Shi","Bai","Qian"};
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    string s;
    cin>>s;
    reverse(s.begin(),s.end());
    int i=s.length()-1,j;
    queue<string> q;
    if(s[i]=='-'){
        i--;
        q.push("Fu");
    }
    if(s[i]=='0'){//特殊情况
        q.push("ling");
        i--;
    }
    for(;i>=0;i--){
        j=i;
        if(i>=0&&s[i]=='0'){
            while(i>=0&&s[i]=='0'){
                i--;
            }
            i++;//保证i一定>=0
            if(i/4==j/4){//0在同一区间
                if(i%4!=0){//连续的0在区间中段
                    q.push("ling");
                }
            }
            else{//多个0跨越区间
                q.push(id[j/4]);//只可能跨越万和个的区间
                if(i!=0){//最后一个0不是个位
                    q.push("ling");
                }
            }
        }
        else{
            q.push(dight[s[i]-'0']);
            if(i%4!=0){//不是个位的情况
                q.push(num[i%4]);
            }
        }
        if(i%4==0&&i>3){//到了区间末尾
            q.push(id[i/4]);
        }
    }
    cout<<q.front();
    q.pop();
    while(!q.empty()){
        cout<<" "<<q.front();
        q.pop();
    }
    cout<<endl;
    return 0;
}