Number reading method (analog)

Description

Professor Tom is to teach a graduate course on genes, one thing made him quite a headache: there are tens of thousands of base pairs on a chromosome, they are numbered from 0 to millions, tens of millions, or billions.
For example, when the students about the bases at position No. 1234567009, the light looks figures are difficult to accurately read out.
So, he desperately needs a system, and then when he entered 1,234,567,009 time, will give the corresponding read the law:
1,234,567,009
represented as Pinyin
shi er yi san qian si bai wu shi liu wan qi qian ling jiu
so he only needs to read shining on it.
Your task is to help him design a system like this: Given a string of digits, you help him into reading and writing Chinese Pinyin string accordance with the norms of the adjacent two syllables with a lattice open spaces.
Note must be in strict accordance with specifications, such as "10010" is read as "yi wan ling yi shi" instead of "yi wan ling shi", " 100000" is read as "shi wan" instead of "yi shi wan", "2000 " read as "er qian" instead of "liang qian".

Input

There is a string of numbers, the numerical size of not more than 2,000,000,000.

Output

Is a string of lower case letters, spaces and commas composition, it represents the number of English reading.

Sample Input

1234567009

Sample Output

shi er yi san qian si bai wu shi liu wan qi qian ling jiu

Meaning of the questions is very simple, patient analysis, simulation and then get away.

Can be found in every four into a group, in the middle billion in Wan connected relatively simple.

Will be divided into good set of four bits of each number separately to determine what

There is a ten shi, yi shi two possibilities based on one hundred cases, such as 1011 are: yi qian ling shi yi, instead of yi qian ling yi shi yi

Note ling and output spaces

Pronunciation uncertainty can go to this website to check online:

http://www.nicetool.net/embed/number_to_chinese.html

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <stack>
  9 #include <queue>
 10 #include <set>
 11 #include <map>
 12 #include <sstream>
 13 const int INF=0x3f3f3f3f;
 14 typedef long long LL;
 15 const int mod=1e9+7;
 16 const int maxn=1e4+10;
 17 using namespace std;
 18  
 19 map<int,string> mp;
 20  
 21 int main()
 22 {
 23     #ifdef DEBUG
 24     freopen("sample.txt","r",stdin);
 25     #endif
 26      
 27     mp[0]="ling";mp[1]="yi"; mp[2]="er"; mp[3]="san"; mp[4]="si"; mp[5]="wu";
 28     mp[6]="liu"; mp[7]="qi"; mp[8]="ba"; mp[9]="JIU " ; MP [ 10 ] = " Shi " ;
 29       
30      int n-;
 31 is      CIN >> n-;
 32      int T, A, B, C, D;
 33 is      int F1 = 0 , F2 = 0 ;
 34 is      IF (n-> = 100000000 ) // process "100 million" before 
35      {
 36          F1 = . 1 ;
 37 [          T = n-/ 100000000 ;
 38 is          A = T / 10 ;
 39          B =% T 10;
 40         if(a)
 41         {
 42             if(a>1) cout<<mp[a]<<" ";
 43             cout<<"shi";
 44             if(b) cout<<" "<<mp[b];
 45         }
 46         else cout<<mp[b];
 47         cout<<" "<<"yi";
 48         if(n%100000000) cout<<" ";// if the latter is not 0, the partition 
49      }
 50      IF (n-% 100000000 > = 10000 ) // process "Wan" Before 
51 is      {
 52 is          F2 = . 1 ;
 53 is          T = (n-% 100000000 ) / 10000 ;
 54 is          A = T / 1000 ;
 55          B =% T 1000 / 100 ;
 56 is          C% T = 100 / 10 ;
 57 is          D% = T 10 ;
 58          IF (A)
 59         {
 60             cout<<mp[a]<<" qian";
 61             if(b||c||d) cout<<" ";
 62         }
 63         else if(f1) cout<<"ling ";
 64         if(b)
 65         {
 66             cout<<mp[b]<<" bai";
 67             if(c||d) cout<<" ";
if71        {
70ELSE69        }
68          
              ((a&&c)||(a&&d)) cout<<"ling ";
 72         }
 73         if(c)
 74         {
 75             if(c>1) cout<<mp[c]<<" ";
 76             cout<<"shi";
 77             if(d) cout<<" ";
 78         }
 79         else
 80         {
 81             if(b&&d) cout<<"ling ";
if83        }
82          (D) << COUT MP [D];
 84          COUT << "  " << " WAN " ;
 85          IF (n-% 10000 ) COUT << "  " ; // if the latter is not 0, the partition 
86      }
 87      IF ( % n- 10000 )
 88      {
 89          T = (n-% 10000 );
 90          A = T / 1000 ;
 91 is          B% T = 1000 / 100 ;
 92          C% T = 100 / 10;
 93         d=t%10;
 94         if(a)
 95         {
 96             cout<<mp[a]<<" qian";
 97             if(b||c||d) cout<<" ";
 98         }
 99         else if(f2) cout<<"ling ";
100         if(b)
101         {
102             cout<<mp[b]<<" bai";
103             if(c||d) cout<<" ";
104         }
105         else
106         {
107             if((a&&c)||(a&&d)) cout<<"ling ";
108         }
109         if(c)
110         {
111             if(c>1) cout<<mp[c]<<" ";
112             cout<<"shi";
113             if(d) cout<<" ";
114         }
115         else
116         {
117             if(b&&d) cout<<"ling ";
118         }
119         if(d) cout<<mp[d];
120     }
121     cout<<endl;
122      
123     return 0;
124 }

I wrote a little trouble, recommended a simple blog writing:

https://blog.csdn.net/weixin_43804406/article/details/104126113