// C++ program to find the largest number that can be   // formed from given sum of digits and number of digits.   #include          using     namespace     std  ;   // Prints the smallest possible number with digit sum 's'   // and 'm' number of digits.   void     findLargest  (  int     m       int     s  )   {      // If sum of digits is 0 then a number is possible      // only if number of digits is 1.      if     (  s     ==     0  )     {      (  m     ==     1  )     ?     cout      < <     'Largest number is '      < <     0      :     cout      < <     'Not possible'  ;      return  ;      }      // Sum greater than the maximum possible sum.      if     (  s     >     9     *     m  )     {      cout      < <     'Not possible'  ;      return  ;      }      // Create an array to store digits of result      int     res  [  m  ];      // Fill from most significant digit to least      // significant digit.      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )     {      // Fill 9 first to make the number largest      if     (  s     >=     9  )     {      res  [  i  ]     =     9  ;      s     -=     9  ;      }      // If remaining sum becomes less than 9 then      // fill the remaining sum      else     {      res  [  i  ]     =     s  ;      s     =     0  ;      }      }      cout      < <     'Largest number is '  ;      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )      cout      < <     res  [  i  ];   }   // Driver code   int     main  ()   {      int     s     =     9       m     =     2  ;      findLargest  (  m       s  );      return     0  ;   }   

   // C program to find the largest number that can be   // formed from given sum of digits and number of digits.   #include         // Prints the smallest possible number with digit sum 's'   // and 'm' number of digits.   void     findLargest  (  int     m       int     s  )   {      // If sum of digits is 0 then a number is possible      // only if number of digits is 1.      if     (  s     ==     0  )     {      (  m     ==     1  )     ?     printf  (  'Largest number is 0'  )      :     printf  (  'Not possible'  );      return  ;      }      // Sum greater than the maximum possible sum.      if     (  s     >     9     *     m  )     {      printf  (  'Not possible'  );      return  ;      }      // Create an array to store digits of result      int     res  [  m  ];      // Fill from most significant digit to least      // significant digit.      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )     {      // Fill 9 first to make the number largest      if     (  s     >=     9  )     {      res  [  i  ]     =     9  ;      s     -=     9  ;      }      // If remaining sum becomes less than 9 then      // fill the remaining sum      else     {      res  [  i  ]     =     s  ;      s     =     0  ;      }      }      printf  (  'Largest number is '  );      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )      printf  (  '%d'       res  [  i  ]);   }   // Driver code   int     main  ()   {      int     s     =     9       m     =     2  ;      findLargest  (  m       s  );      return     0  ;   }   // This code is contributed by Sania Kumari Gupta   

   // Java program to find the largest number that can be   // formed from given sum of digits and number of digits   class   GFG      {      // Function to print the largest possible number with digit sum 's'      // and 'm' number of digits      static     void     findLargest  (  int     m       int     s  )      {      // If sum of digits is 0 then a number is possible      // only if number of digits is 1      if     (  s     ==     0  )      {      System  .  out  .  print  (  m     ==     1     ?     'Largest number is 0'     :     'Not possible'  );             return     ;      }          // Sum greater than the maximum possible sum      if     (  s     >     9  *  m  )      {      System  .  out  .  println  (  'Not possible'  );      return     ;      }          // Create an array to store digits of result      int  []     res     =     new     int  [  m  ]  ;          // Fill from most significant digit to least      // significant digit      for     (  int     i  =  0  ;     i   <  m  ;     i  ++  )      {      // Fill 9 first to make the number largest      if     (  s     >=     9  )      {      res  [  i  ]     =     9  ;      s     -=     9  ;      }          // If remaining sum becomes less than 9 then      // fill the remaining sum      else      {      res  [  i  ]     =     s  ;      s     =     0  ;      }      }          System  .  out  .  print  (  'Largest number is '  );      for     (  int     i  =  0  ;     i   <  m  ;     i  ++  )      System  .  out  .  print  (  res  [  i  ]  );      }          // driver program      public     static     void     main     (  String  []     args  )         {      int     s     =     9       m     =     2  ;      findLargest  (  m       s  );      }   }   // Contributed by Pramod Kumar   

   # Python 3 program to find   # the largest number that    # can be formed from given   # sum of digits and number   # of digits.   # Prints the smallest    # possible number with digit    # sum 's' and 'm' number of   # digits.   def   findLargest  (   m     s  )   :   # If sum of digits is 0   # then a number is possible   # only if number of digits   # is 1.   if   (  s   ==   0  )   :   if  (  m   ==   1  )   :   print  (  'Largest number is '      '0'    end   =   ''  )   else   :   print  (  'Not possible'    end   =   ''  )   return   # Sum greater than the    # maximum possible sum.   if   (  s   >   9   *   m  )   :   print  (  'Not possible'    end   =   ''  )   return   # Create an array to    # store digits of   # result   res   =   [  0  ]   *   m   # Fill from most significant   # digit to least significant   # digit.   for   i   in   range  (  0     m  )   :   # Fill 9 first to make   # the number largest   if   (  s   >=   9  )   :   res  [  i  ]   =   9   s   =   s   -   9   # If remaining sum    # becomes less than    # 9 then fill the    # remaining sum   else   :   res  [  i  ]   =   s   s   =   0   print  (   'Largest number is '    end   =   ''  )   for   i   in   range  (  0     m  )   :   print  (  res  [  i  ]  end   =   ''  )   # Driver code   s   =   9   m   =   2   findLargest  (  m     s  )   # This code is contributed by Nikita Tiwari.   

   // C# program to find the    // largest number that can    // be formed from given sum    // of digits and number of digits   using     System  ;   class     GFG   {          // Function to print the       // largest possible number       // with digit sum 's' and       // 'm' number of digits      static     void     findLargest  (  int     m       int     s  )      {      // If sum of digits is 0       // then a number is possible       // only if number of digits is 1      if     (  s     ==     0  )      {      Console  .  Write  (  m     ==     1     ?         'Largest number is 0'     :         'Not possible'  );             return     ;      }      // Sum greater than the      // maximum possible sum      if     (  s     >     9     *     m  )      {      Console  .  WriteLine  (  'Not possible'  );      return     ;      }      // Create an array to       // store digits of result      int     []  res     =     new     int  [  m  ];      // Fill from most significant       // digit to least significant digit      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )      {      // Fill 9 first to make      // the number largest      if     (  s     >=     9  )      {      res  [  i  ]     =     9  ;      s     -=     9  ;      }      // If remaining sum becomes       // less than 9 then      // fill the remaining sum      else      {      res  [  i  ]     =     s  ;      s     =     0  ;      }      }      Console  .  Write  (  'Largest number is '  );      for     (  int     i     =     0  ;     i      <     m  ;     i  ++  )      Console  .  Write  (  res  [  i  ]);      }          // Driver Code      static     public     void     Main     ()      {      int     s     =     9       m     =     2  ;      findLargest  (  m       s  );      }   }   // This code is Contributed by ajit   

      // PHP program to find the largest    // number that can be formed from    // given sum of digits and number    // of digits.   // Prints the smallest possible    // number with digit sum 's'    // and 'm' number of digits.   function   findLargest  (  $m     $s  )   {   // If sum of digits is 0 then    // a number is possible only if   // number of digits is 1.   if   (  $s   ==   0  )   {   if  ((  $m   ==   1  )   ==   true  )   echo   'Largest number is '      0  ;   else   echo   'Not possible'  ;   return   ;   }   // Sum greater than the   // maximum possible sum.   if   (  $s   >   9   *   $m  )   {   echo   'Not possible'  ;   return   ;   }   // Create an array to store    // digits of result Fill from    // most significant digit to    // least significant digit.   for   (  $i   =   0  ;   $i    <   $m  ;   $i  ++  )   {   // Fill 9 first to make   // the number largest   if   (  $s   >=   9  )   {   $res  [  $i  ]   =   9  ;   $s   -=   9  ;   }   // If remaining sum becomes    // less than 9 then fill    // the remaining sum   else   {   $res  [  $i  ]   =   $s  ;   $s   =   0  ;   }   }   echo   'Largest number is '  ;   for   (  $i   =   0  ;   $i    <   $m  ;   $i  ++  )   echo   $res  [  $i  ];   }   // Driver code   $s   =   9  ;   $m   =   2  ;   findLargest  (  $m     $s  );   // This code is contributed by m_kit    ?>   

    <  script  >   // Javascript program to find the largest number that can be   // formed from given sum of digits and number of digits.   // Prints the smallest possible number with digit sum 's'   // and 'm' number of digits.   function     findLargest  (  m       s  )   {      // If sum of digits is 0 then a number is possible      // only if number of digits is 1.      if     (  s     ==     0  )      {      (  m     ==     1  )  ?     document  .  write  (  'Largest number is '     +     0  )      :     document  .  write  (  'Not possible'  );      return     ;      }      // Sum greater than the maximum possible sum.      if     (  s     >     9  *  m  )      {      document  .  write  (  'Not possible'  );      return     ;      }      // Create an array to store digits of result      let     res     =     new     Array  (  m  );      // Fill from most significant digit to least      // significant digit.      for     (  let     i  =  0  ;     i   <  m  ;     i  ++  )      {      // Fill 9 first to make the number largest      if     (  s     >=     9  )      {      res  [  i  ]     =     9  ;      s     -=     9  ;      }      // If remaining sum becomes less than 9 then      // fill the remaining sum      else      {      res  [  i  ]     =     s  ;      s     =     0  ;      }      }      document  .  write  (  'Largest number is '  );      for     (  let     i  =  0  ;     i   <  m  ;     i  ++  )      document  .  write  (  res  [  i  ]);   }   // Driver code      let     s     =     9       m     =     2  ;      findLargest  (  m       s  );   // This code is contributed by Mayank Tyagi    <  /script>   

   #include          #include         using     namespace     std  ;   int     largest_number  (  int     s       int     d  )     {      if     (  s     ==     0  )     {      return     0  ;      }      if     (  s     >     9     *     d  )     {      return     -1  ;      }      string     result     =     ''  ;      for     (  int     i     =     0  ;     i      <     d  ;     i  ++  )     {      if     (  s     >=     9  )     {      result     +=     '9'  ;      s     -=     9  ;      }     else     {      result     +=     to_string  (  s  );      s     =     0  ;      }      if     (  s     ==     0     &&     i      <     d  -1  )     {      result     +=     string  (  d  -  i  -1       '0'  );      break  ;      }      }      return     stoi  (  result  );   }   int     main  ()     {      // Test case 1      cout      < <     largest_number  (  9       2  )      < <     endl  ;     // Output: 90      // Test case 2      cout      < <     largest_number  (  20       3  )      < <     endl  ;     // Output: 992      return     0  ;   }   

   import     java.util.*  ;   public     class   Main     {      public     static     int     largest_number  (  int     s       int     d  )      {      // If s is 0 then the largest number is 0.      if     (  s     ==     0  )     {      return     0  ;      }      // If s is greater than 9 times d then it is      // impossible to form a d-digit number whose sum of      // digits is s.      if     (  s     >     9     *     d  )     {      return     -  1  ;      }      // Initialize an empty string to store the result.      String     result     =     ''  ;      // Loop through each digit of the number.      for     (  int     i     =     0  ;     i      <     d  ;     i  ++  )     {      // If s is greater than or equal to 9 then add      // 9 to the result and subtract 9 from s.      if     (  s     >=     9  )     {      result     +=     '9'  ;      s     -=     9  ;      }      // Otherwise add s to the result and set s to      // 0.      else     {      result     +=     Integer  .  toString  (  s  );      s     =     0  ;      }      // If s is 0 and there are still digits left to      // fill then fill the remaining digits with 0s      // and break out of the loop.      if     (  s     ==     0     &&     i      <     d     -     1  )     {      result     +=     String  .  join  (      ''        Collections  .  nCopies  (  d     -     i     -     1       '0'  ));      break  ;      }      }      // Convert the result to an integer and return it.      return     Integer  .  parseInt  (  result  );      }      public     static     void     main  (  String  []     args  )      {      // Test case 1      System  .  out  .  println  (      largest_number  (  9       2  ));     // Output: 90      // Test case 2      System  .  out  .  println  (      largest_number  (  20       3  ));     // Output: 992      }   }   

   def   largest_number  (  s     d  ):   if   s   ==   0  :   return   0   if   s   >   9   *   d  :   return   -  1   result   =   ''   for   i   in   range  (  d  ):   if   s   >=   9  :   result   +=   '9'   s   -=   9   else  :   result   +=   str  (  s  )   s   =   0   if   s   ==   0   and   i    <   d  -  1  :   result   +=   '0'   *   (  d  -  i  -  1  )   break   return   int  (  result  )   # Test case 1   print  (  largest_number  (  9     2  ))   # Output: 90   # Test case 2   print  (  largest_number  (  20     3  ))   # Output: 992   

   using     System  ;   class     Program     {      static     int     LargestNumber  (  int     s       int     d  )     {      if     (  s     ==     0  )     {      return     0  ;      }      if     (  s     >     9     *     d  )     {      return     -  1  ;      }      string     result     =     ''  ;      for     (  int     i     =     0  ;     i      <     d  ;     i  ++  )     {      if     (  s     >=     9  )     {      result     +=     '9'  ;      s     -=     9  ;      }     else     {      result     +=     s  .  ToString  ();      s     =     0  ;      }      if     (  s     ==     0     &&     i      <     d     -     1  )     {      result     +=     new     string  (  '0'       d     -     i     -     1  );      break  ;      }      }      return     int  .  Parse  (  result  );      }      static     void     Main  (  string  []     args  )     {      // Test case 1      Console  .  WriteLine  (  LargestNumber  (  9       2  ));     // Output: 90      // Test case 2      Console  .  WriteLine  (  LargestNumber  (  20       3  ));     // Output: 992      }   }   

   function     largestNumber  (  s       d  )     {   if     (  s     ==     0  )     {   return     0  ;   }   if     (  s     >     9     *     d  )     {   return     -  1  ;   }   let     result     =     ''  ;   for     (  let     i     =     0  ;     i      <     d  ;     i  ++  )     {   if     (  s     >=     9  )     {   result     +=     '9'  ;   s     -=     9  ;   }     else     {   result     +=     s  .  toString  ();   s     =     0  ;   }   if     (  s     ==     0     &&     i      <     d     -     1  )     {   result     +=     '0'  .  repeat  (  d     -     i     -     1  );   break  ;   }   }   return     parseInt  (  result  );   }   // Test cases   console  .  log  (  largestNumber  (  9       2  ));     // Output: 90   console  .  log  (  largestNumber  (  20       3  ));     // Output: 992