asp.net conversion between literal and integer characters in C #

ASP tutorials conversion between characters and integers in. NET C #
Digital to Chinese basic ideas:

is to decompose

1: Computation of the most basic [0,9999] digital Translations

2: Through the ' million ' word to connect two adjacent to the (1) calculated results, get [0,99999999] scope of translation.

3: Through the ' billion ' word connection, get a large number of translations within the range.

The basic idea of Chinese digital conversion:

Main question: 1: Where do I multiply? 2: What is the scope of multiplication?

Easy to enter the misunderstanding:

1: Only consider where to multiply, such as 10 million

(It is easy to think that, when the word "million" is calculated, the nearest unit is ' thousand ', in which case the multiplication is performed, and in addition to the multiplication, the order is computed).

* * Test data: 130 million. The expression of the algorithm is: 1e8 + 3000 * 10000. Since it is a sequential calculation, the actual calculation is (1E8 + 3000) * 10000

Got the wrong answer.

2: Taking into account the problems encountered in (1), but the multiplication of the scope of judgment error. such as data 130 million, the correct expression is 1e8 + (3000 * 10000)

It is assumed that the ' * ' and ' + ' priority is taken into account in the calculation of expressions that are similar to only ' + ' and ' * '.

such as "130 million" can get the correct answer.

* * Test data: 101.2 million. The algorithm by (2) is: 1e 8 + 100 + (20 * 10000).

and the actual result should be: 1e8 + (100 + 20) * 10000, got the wrong answer.

summed up the solution: that is to consider 1: where to multiply? 2: What is the scope of multiplication?

* * such as 101.2 million. When you encounter ' million ', the previous ' hundred ' < ' million ' determines the current need for multiplication.

* * Again by ' million ' before the first > ' Million ' units (here for billion), determines the multiplication of the scope of the two between.

Get the correct expression of the translation: 1e8 + (100 + 20) * 10000

Private Class Num2chinese
{
private static readonly string unit = "Chichong";
private static readonly string ala = "0123456789";
private static readonly String ch = "0123456789";

private static string Transsmall (String num)


{/*[0, 10,000) translation of the interval * *


String str = "";


String nn = Num.trimstart (' 0 ');


BOOL flag = FALSE;


int j = 0;


for (int i = nn.length-1 i &gt;= 0; I--, j + +)


{


int no = Ala.indexof (Nn[i]);


if (no!= 0) flag = true;


if (flag)


{


if (J!= 0 &amp;&amp; no!= 0)


str = unit[j];


if (!) ( No = 0 &amp;&amp; str[str.length-1] = = ' 0 ')


str = Ch[no];


}


}


Char[] s = Str.tochararray ();


Array.reverse (s);


str = new string (s);


return str;


}

private static string trans (string num, int range, char ss)
{/* recursive solution. [0, Thousand] through the word "million" can be spliced into [0, tens of millions], the same as [0, tens of millions] through the billion word connection.
Parameter meaning:
The NUM string is decomposed as a group per range and is connected by the SS character. (4-digit group or eight-digit group) * *
string ret = "";
string input = Num.trimstart (' 0 ');

for (int i = Input.length-range i &gt;-range i-= range)


{


int st = i, Len = range;/*st and Len respectively indicate the starting position and length of each set of strings * *


if (I &lt; 0)


{


st = 0;


Len = range + i;


}


string tmp = input.substring (St, Len);


Long nn = long.parse (TMP);


String cur = "";


if (NN = 0)


{


RET = ss + ret;


Continue


}


if (ss = ' million ')


{


cur = transsmall (TMP) + SS;


}


Else


{


cur = trans (tmp, 4, ' million ') + SS;


}/* calculate the value of each group, stitching SS characters to the end.


if (nn% = = 0 &amp;&amp; ret.length &gt; 0 &amp;&amp; ret[0]!= ' 0 ')


{/* The current group at the end of zero to determine whether + 0 * *


int LL = Num.length-st-len;


LL = ll &gt; 8? 8:ll;/


/* The largest unit is billion, then the current group after 8-bit range without zero output 0


Example: 100,000,005, 0 requires output, 100,000,005 (not 100,000,005) * *


String tt = num.substring (st + Len, LL);


if (Tt.trim (' 0 ')!= "")//after eight digits there is a non-0 value


Cur + = ' 0 ';


}


if (ss = ' million ' &amp;&amp; nn &lt; 1000 | |


ss = ' billion ' &amp;&amp; nn &lt; (long) 1e7)


{/* Before the current group position is vacant, the front position is 0 * * *


cur = ' 0 ' + cur;


}


RET = cur + ret;


}

int s = 0, _len = ret.length-1;


if (ret[0] = = ' 0 ')


{


s = 1;


_len--;


}


if (ret[ret.length-2] = = ' 0 ')


{


_len--;


}/* minus the extra front and back bits of the SS-unit character after the substring * *


ret = ret.substring (s, _len);


return ret;


}

public static string Transbiginteger (String ss)


{/* call trans for translation and exception handling, support large number * *


string ret = Ss.trimstart (' 0 ');


String tt = "0123456789";


char[] ch = tt.tochararray ();


ret = Ret.trim (');


if (Ret.trim (CH)!= "")


{


MessageBox.Show ("Enter data illegal!") ");


Return "";


}


if (ret = "")


{


return "0";


}


ret = trans (ret, 8, ' billion ');


return ret;


}


}

Private Class Chinesenumconvert
{/* integer to numeric class, does not support large number
Key point: Where do you multiply? Calculate the range of multiplication? */
private static readonly string digit = "0123456789";
private static readonly string unit = "Chivan";
private static readonly int[] num = {1, 10, 100, 1000, 10000, 100000000};

private static bool Checkstring (string word)


{/* Simple to determine whether the input is legal 1; other characters. 2: There are not 0 digits adjacent to the * *


Word = Word.trim (');


String tt = digit + unit;


string tmp = Word.trim (Tt.tochararray ());


if (tmp!= "")


return false;


for (int i = 0; i &lt; word.length-1; i++)


{


if (Digit.indexof (Word[i]) &gt;= 1


&amp;&amp; Digit.indexof (word[i + 1]) &gt;= 1)


{


return false;


}


}


return true;


}

         public static long Chinesenum2int (string word)
          {
              if (!checkstring (word))
             {
                 MessageBox.Show ("Enter data illegal!") ");
                 return 0;
            }
             Long sum = 0;
             Long tmp = 0;
             long[] tt = new long[150];

int dig =-1, unit0 =-1, unit1 =-1, unit2 =-1;


if (Unit.indexof (word[0]) &gt;= 0)


{


Word = "one" + word;


}


if (Unit.indexof (word[word.length-1]) &lt; 0)


{


if (Word.length &gt;= 2 &amp;&amp; (tmp = Unit.indexof (word[word.length-2))) &gt; 0)


{


Word = Word + unit.tochararray () [tmp-1];


}


Else


{


Word = = ' a ';


}


}


for (int i = 1; i &lt; word.length; i++)


{


Dig = Digit.indexof (word[i-1]);


Unit0 = Unit.indexof (Word[i]);

if (unit0 &gt;= 0)


{


Unit1 =-1; Unit2 =-1;


int idx =-1;


for (int j = i-1 J &gt;= 0; j--)


{


Unit1 = Unit.indexof (Word[j]);


if (unit1 &gt;= 0) break;


}//Whether you need to do multiplication


for (int j = i-1 J &gt;= 0; j--)


{


Unit2 = Unit.indexof (Word[j]);


if (Unit2 &gt; Unit0)


{


idx = j;


Break


}


}//determines the scope of multiplication


if (unit1 &gt;= 0 &amp;&amp; unit0 &gt;= unit1)


{


if (Dig &gt;= 0) sum + = dig;


if (unit2 = = 1)


{


Sum *= num[unit0];


}


Else


{


sum = (Sum-tt[idx]) * Num[unit0] + TT[IDX];


}


}


Else


{


if (Dig &lt; 0) dig = 1;


Sum + + dig * num[unit0];


}


}


Tt[i] = sum;


}


return sum;


}