Imitation NetEase one-dollar capture treasure algorithm, wide-length data calculation, string implementation

/**

* @company

* @version 1.0

* @author fangjianming

* @email [email protected]

* @date December 10, 2014

*/

public class Cal {

public static void Main (string[] args) {

String value = "581999974";

String value2 = "46658";

Long bet = 100;

String value3 = "10000001";

//Calculate sum

int []eleven = addstring (value,value2);

//calculation to find remainder

Long result = Divideint (Eleven,bet);

//Calculation results

int []intvalue = addstring (value3,string.valueof (result));

//int[] turn char[]

Char [] charvalue = Transintarraytochararray (intvalue);

System.out.print (charvalue);

}

//int[] turn char[]

private static char[] Transintarraytochararray (int[] intvalue) {

if (intvalue = = NULL | | intvalue.length ==0)

throw new NullPointerException ("Eleven must bot be null");

char[] charvalue = new Char[intvalue.length];

int i = 0;

for (int value:intvalue)

{

charvalue[i++] = (char) (value +48);

}

return charvalue;

}

//calculation to find remainder

private static Long Divideint (int[] Eleven, long divide) {

if (eleven = = NULL | | eleven.length ==0)

throw new NullPointerException ("Eleven must bot be null");

if (divide = = 0)

throw new NullPointerException ("Divide must bot be zero");

Long result = 0;

int len = eleven.length;

for (int i = 0;i<len; i++)

{

int tmp = eleven[i];

result = Tmp+10*result;

if (result >= divide)

{

result%= divide;

}

}

return result;

}

//Calculate sum

public static int [] AddString (String value1,string value2)

{

if (value1 = = NULL | | value2 = = NULL | | value1 = = "" "| | value2 = = "")

throw new NullPointerException ("value1 and value2 must bot be null");

String max = (value1.length () > Value2.length ())? Value1:value2;

String min = (value1.length () <= value2.length ())? Value1:value2;

int len = Max.length ();

int len2 = Min.length ();

int ten = 0;

int ge = 0;

int []eleven = new Int[len];

for (int i = max.length () -1,j = Min.length ()-1; I >= 0; I--, j--)

{

int a1 = 0;

int a2 = 0;

a1 = Max.charat (i)-48;

if (j>=0)

A2 = Min.charat (I-(Max.length ()-min.length ()))-48;

ge = (a1+a2+ten)% 10;

ten = (A1+A2)/10;

eleven[i] = GE;

}

if (Ten >0)

{

int []eleven2 = new int[len+1];

eleven2[0] = ten;

for (int j = 0;j <len; j + +)

{

eleven2[j+1] = eleven[j];

}

return eleven2;

}

return eleven;

}

}

Imitation NetEase one-dollar capture treasure algorithm, wide-length data calculation, string implementation