Un nombre n i una base negativa negBase se'ns dóna que hem de representar n en aquesta base negativa. La base negativa funciona semblant a la base positiva. Per exemple, a la base 2 multipliquem els bits a 1 2 4 8 i així successivament per obtenir el nombre real en decimal. En el cas de la base -2 hem de multiplicar els bits amb 1 -2 4 -8 i així successivament per obtenir el nombre en decimal.
Exemples:
recorregut posterior a la comanda
Input : n = 13 negBase = -2 Output : 11101 1*(16) + 1*(-8) + 1*(4) + 0*(-2) + 1*(1) = 13
És possible representar un nombre en qualsevol base negativa amb el mateix procediment (Consulteu Una setmana per als detalls). Per simplificar (per desfer-se dels caràcters A B, etc. a la sortida) estem permetent que la nostra base només estigui entre -2 i -10.
Podem resoldre aquest problema de manera semblant a la resolució d'un problema amb bases positives, però una cosa important a recordar és que la resta sempre serà positiva tant si treballem amb una base positiva com amb una base negativa, però en la majoria de compiladors el resultat de dividir un nombre negatiu per un nombre negatiu s'arrodoneix cap a 0, normalment deixant un residu negatiu.
Així, sempre que aconseguim un residu negatiu, el podem convertir en positiu com a continuació
Let n = (?negBase) * quotient + remainder = (?negBase) * quotient + negBase ? negBase + negBase = (?negBase) * (quotient + 1) + (remainder + negBase). So if after doing 'remainder = n % negBase' and 'n = n/negBase' we get negative remainder we do following. remainder = remainder + (-negBase) n = n + 1 Example : n = -4 negBase = -3 In C++ we get remainder = n % negBase = -4/-3 = -1 n = n/negBase [Next step for base conversion] = -4/-3 = 1 To avoid negative remainder we do remainder = -1 + (-negBase) = -1 - (-3) = 2 n = n + 1 = 1 + 1 = 2.
Així, quan tinguem un residu negatiu, el farem positiu afegint-li el valor absolut de la base i afegint 1 al nostre quocient.
L'enfocament explicat anteriorment s'implementa al codi següent
quantes ciutats hi ha als EUAC++
// C/C++ program to convert n into negative base form #include
// Java program to convert n into // negative base form class GFG { // Method to convert n to base negBase static String toNegativeBase(int n int negBase) { // If n is zero then in any base // it will be 0 only if (n == 0) return '0'; String converted = ''; while (n != 0) { // Get remainder by negative base // it can be negative also int remainder = n % negBase; n /= negBase; // if remainder is negative // add Math.abs(base) to it // and add 1 to n if (remainder < 0) { remainder += (-negBase); n += 1; } // convert remainder to String add into the result converted = String.valueOf(remainder) + converted; } return converted; } // Driver Code public static void main(String[] args) { int n = 13; int negBase = -2; System.out.print(toNegativeBase(n negBase)); } } // This code is contributed by 29AjayKumar
# Python 3 program to convert n into # negative base form # Method to convert n to base negBase def toNegativeBase(n negBase): # If n is zero then in any base it # will be 0 only if (n == 0): return '0' converted = '01' while (n != 0): # Get remainder by negative base # it can be negative also remainder = n % (negBase) n = int(n/negBase) # if remainder is negative add # abs(base) to it and add 1 to n if (remainder < 0): remainder += ((-1) * negBase) n += 1 # convert remainder to string add # into the result converted = str(remainder) + converted return converted # Driver Code if __name__ == '__main__': n = 13 negBase = -2 print(toNegativeBase(n negBase)) # This code is contributed by # Surendra_Gangwar
// C# program to convert n into // negative base form using System; class GFG { // Method to convert n to base negBase static String toNegativeBase(int n int negBase) { // If n is zero then in any base // it will be 0 only if (n == 0) return '0'; String converted = ''; while (n != 0) { // Get remainder by negative base // it can be negative also int remainder = n % negBase; n /= negBase; // if remainder is negative // add Math.Abs(base) to it // and add 1 to n if (remainder < 0) { remainder += (-negBase); n += 1; } // convert remainder to String add into the result converted = String.Join('' remainder) + converted; } return converted; } // Driver Code public static void Main(String[] args) { int n = 13; int negBase = -2; Console.Write(toNegativeBase(n negBase)); } } // This code is contributed by Rajput-Ji
<script> // JavaScript program to convert n into // negative base form // Method to convert n to base negBase function toNegativeBase(n negBase) { // If n is zero then in any base // it will be 0 only if (n == 0) return '0'; let converted = '01'; while (n != 0) { // Get remainder by negative base // it can be negative also let remainder = (-1)*(Math.abs(n) % Math.abs(negBase)); n = parseInt(n/negBase); // if remainder is negative // add Math.abs(base) to it // and add 1 to n if (remainder < 0) { remainder += ((-1)*negBase); n += 1; } // convert remainder to String add into the result converted = remainder.toString() + converted; } return converted; } // Driver Code let n = 13; let negBase = -2; document.write(toNegativeBase(n negBase)''); // This code is contributed by shinjanpatra </script>
Sortida:
11101
Complexitat temporal: O(N)
Espai auxiliar: O(1)
Referència:
https://en.wikipedia.org/wiki/Negative_base