Common Library of functions (collected in Network)

Large number multiplication: void _mult (String num1, String num2, String &result) Parameters 1 and 2 are two large numbers, the result is saved in the 3rd string

1 void_mult (stringNUM1,stringNUM2,string&result)2 {3Reverse (Num1.begin (), Num1.end ());//reversal4 Reverse (Num2.begin (), Num2.end ());5result="";6     intI, J, re_int[ Max];//******** Here 150 is the number of digits that can be increased or decreased as needed *********7memset (Re_int,0,sizeof(Re_int));8      for(i=0; I<num1.length (); i++)//two string multiplication, the result is stored in the Re_int array, up to 150 bits! 9          for(j=0; J<num2.length (); J + +)TenRE_INT[I+J] + = ((num1[i]- -) * (num2[j]- -));  One     intJinwei=0, Zhi; A      for(i=0; I<num1.length () +num2.length (); i++)//handling the carry problem individually, the array in the previous step can have more than 10 of each element, so no rounding is handled -     { -Zhi = re_int[i]+Jinwei; theRe_int[i] = zhi%Ten; -Jinwei = zhi/Ten; -     } -      for(I=num1.length () +num2.length ()-1; i>=0; i--)//Mark I with a marker, the front part of the array re_int may be 0, to remove +         if(re_int[i]!=0) Break;  -      for(; i>=0; i--)//Convert an integer array to a string +result = result+ (Char) (re_int[i]+ -); A     if(result=="")//If the result is still empty, the result of multiplication is 0?  atresult="0";  -}

Multiply

Large number of additions: void _plus (String num1,string num2,string &result) Parameters 1 and 2 are two large numbers, the result is saved in the 3rd string

1 void_plus (stringNUM1,stringNUM2,string&result)2 {3result="";4 Reverse (Num1.begin (), Num1.end ());5 Reverse (Num2.begin (), Num2.end ());6     inti;7      for(i=0;i<int(Num1.length ()) &&i<int(Num2.length ()); i++)8     {9         CharC= (Char) (num1[i]+num2[i]- -);Tenresult=result+C; One     } A      while(i<int(Num1.length ())) -     { -result=result+Num1[i]; thei++; -     } -      while(i<int(Num2.length ())) -     { +result=result+Num2[i]; -i++; +     } A     intJinwei=0; at      for(i=0;i<int(Result.length ()); i++) -     { -         intzhi=result[i]- -+Jinwei; -Result[i]= (Char) (zhi%Ten+ -); -jinwei=zhi/Ten; -     } in     if(jinwei!=0) -result = result+ (Char) (jinwei+ -); to      for(I=result.length ()-1; i>=0; i--) +         if(result[i]!='0') Break; -result = Result.substr (0, i+1); the Reverse (Result.begin (), Result.end ()); *     if(result.length () = =0) $result="0";Panax Notoginseng}

Plus

Common Library of functions (collected in Network)