Če je podano binarno število kot niz, izpišite njegove komplemente 1 in 2.

1 je dopolnilo binarnega števila je drugo binarno število, pridobljeno s preklapljanjem vseh bitov v njem, tj. s pretvorbo bita 0 v 1 in bita 1 v 0. V formatu komplementa 1 ostanejo pozitivna števila nespremenjena. Negativna števila dobimo tako, da vzamemo komplement pozitivnih dvojnikov 1.

na primer +9 bo predstavljeno kot 00001001 v osembitnem zapisu, -9 pa bo predstavljeno kot 11110110, kar je komplement 1 za 00001001.

Primeri:

1's complement of '0111' is '1000' 1's complement of '1100' is '0011' 

2 dopolnilo binarnega števila je 1, dodano k komplementu 1 binarnega števila. V predstavitvi binarnih števil z komplementom 2 MSB predstavlja znak z '0', ki se uporablja za znak plus, in '1', ki se uporablja za znak minus. preostali biti se uporabljajo za predstavitev velikosti. pozitivne magnitude so predstavljene na enak način kot v primeru predznakovnega bita ali komplementarne predstavitve 1. Negativne magnitude so predstavljene s komplementom 2 svojih pozitivnih dvojnikov.

Primeri:

2's complement of '0111' is '1001' 2's complement of '1100' is '0100' 

Še en trik za iskanje komplementa dveh:

Korak 1: Začnite od najmanj pomembnega bita in se premikajte levo, dokler ne najdete 1. Dokler ne najdete 1, ostanejo bitovi enaki

2. korak: Ko ste našli 1, pustite 1, kot je, in zdaj

3. korak: Obrnite vse preostale dele v 1.

Ilustracija

Recimo, da moramo najti 2s komplement od 100100

Korak 1: Premikajte in pustite bit enak, dokler ne najdete 1. Tukaj x še ni znan. Odgovor = xxxx00 –

2. korak : Našli ste 1. Naj ostane tako. Odgovor = xxx100

3. korak: Obrnite vse preostale bite v 1. Odgovor = 011100.

Zato je komplement 2s od 100100 011100.

Dopolnitev priporočene prakse 1. Poskusite!

Za svoje dopolnilo moramo preprosto obrniti vse bite.
Za komplement 2 najprej poiščemo komplement ena. Prečkamo komplement enice, začenši z LSB (najmanj pomemben bit), in iščemo 0. Obračamo vse 1 (spremenimo v 0), dokler ne najdemo 0. Na koncu obrnemo najdeno 0. Na primer, komplement 2 za 01000 je 11000 (Upoštevajte, da najprej najdemo svoj komplement 01000 kot 10111). Če so vse 1 (v enem komplementu), v nizu dodamo dodatno 1. Na primer, komplement 2 za 000 je 1000 (komplement 1 za 000 je 111).

Spodaj je izvedba.

C++

// C++ program to print 1's and 2's complement of> // a binary number> #include> using> namespace> std;> > // Returns '0' for '1' and '1' for '0'> char> flip(> char> c) {> return> (c ==> '0'> )?> '1'> :> '0'> ;}> > // Print 1's and 2's complement of binary number> // represented by 'bin'> void> printOneAndTwosComplement(string bin)> {> > int> n = bin.length();> > int> i;> > > string ones, twos;> > ones = twos => ''> ;> > > // for ones complement flip every bit> > for> (i = 0; i ones += flip(bin[i]); // for two's complement go from right to left in // ones complement and if we get 1 make, we make // them 0 and keep going left when we get first // 0, make that 1 and go out of loop twos = ones; for (i = n - 1; i>= 0; i--) { if (ones[i] == '1') twos[i] = '0'; else { twos[i] = '1'; odmor; } } // Če ni prekinitve: vsi so 1 kot v 111 ali 11111; // v takem primeru dodajte dodatno 1 na začetku if (i == -1) twos = '1' + twos; cout < < '1's complement: ' < < ones < < endl; cout < < '2's complement: ' < < twos < < endl; } // Driver program int main() { string bin = '1100'; printOneAndTwosComplement(bin); return 0; }>

Java

// Java program to print 1's and 2's complement of> // a binary number> > class> GFG> {> > > // Returns '0' for '1' and '1' for '0'> > static> char> flip(> char> c)> > {> > return> (c ==> '0'> ) ?> '1'> :> '0'> ;> > }> > > // Print 1's and 2's complement of binary number> > // represented by 'bin'> > static> void> printOneAndTwosComplement(String bin)> > {> > int> n = bin.length();> > int> i;> > > String ones => ''> , twos => ''> ;> > ones = twos => ''> ;> > > // for ones complement flip every bit> > for> (i => 0> ; i { ones += flip(bin.charAt(i)); } // for two's complement go from right to left in // ones complement and if we get 1 make, we make // them 0 and keep going left when we get first // 0, make that 1 and go out of loop twos = ones; for (i = n - 1; i>= 0; i--) { if (ones.charAt(i) == '1') { twos = twos.substring(0, i) + '0' + twos.substring(i + 1); } else { twos = twos.substring(0, i) + '1' + twos.substring(i + 1); odmor; } } // Če ni prekinitve: vsi so 1 kot v 111 ali 11111; // v takem primeru dodajte dodatno 1 na začetku if (i == -1) { twos = '1' + twos; } System.out.println(komplement '1: ' + one);; System.out.println('2's komplement: ' + dvojke); } // Koda gonilnika public static void main(String[] args) { String bin = '1100'; printOneAndTwosComplement(bin); } } // To kodo je prispeval Rajput-Ji>

Python3

# Python3 program to print 1's and 2's> # complement of a binary number> > # Returns '0' for '1' and '1' for '0'> def> flip(c):> > return> '1'> if> (c> => => '0'> )> else> '0'> > # Print 1's and 2's complement of> # binary number represented by 'bin'> def> printOneAndTwosComplement(> bin> ):> > > n> => len> (> bin> )> > ones> => ''> > twos> => ''> > > # for ones complement flip every bit> > for> i> in> range> (n):> > ones> +> => flip(> bin> [i])> > > # for two's complement go from right> > # to left in ones complement and if> > # we get 1 make, we make them 0 and> > # keep going left when we get first> > # 0, make that 1 and go out of loop> > ones> => list> (ones.strip(''))> > twos> => list> (ones)> > for> i> in> range> (n> -> 1> ,> -> 1> ,> -> 1> ):> > > if> (ones[i]> => => '1'> ):> > twos[i]> => '0'> > else> :> > twos[i]> => '1'> > break> > > i> -> => 1> > # If No break : all are 1 as in 111 or 11111> > # in such case, add extra 1 at beginning> > if> (i> => => -> 1> ):> > twos.insert(> 0> ,> '1'> )> > > print> (> '1's complement: '> ,> *> ones, sep> => '')> > print> (> '2's complement: '> ,> *> twos, sep> => '')> > # Driver Code> if> __name__> => => '__main__'> :> > bin> => '1100'> > printOneAndTwosComplement(> bin> .strip(''))> > # This code is contributed> # by SHUBHAMSINGH10>

C#

// C# program to print 1's and 2's complement of> // a binary number> using> System;> > class> GFG> {> > > // Returns '0' for '1' and '1' for '0'> > static> char> flip(> char> c)> > {> > return> (c ==> '0'> ) ?> '1'> :> '0'> ;> > }> > > // Print 1's and 2's complement of binary number> > // represented by 'bin'> > static> void> printOneAndTwosComplement(String bin)> > {> > int> n = bin.Length;> > int> i;> > > String ones => ''> , twos => ''> ;> > ones = twos => ''> ;> > > // for ones complement flip every bit> > for> (i = 0; i { ones += flip(bin[i]); } // for two's complement go from right to left in // ones complement and if we get 1 make, we make // them 0 and keep going left when we get first // 0, make that 1 and go out of loop twos = ones; for (i = n - 1; i>= 0; i--) { if (enice[i] == '1') { twos = twos.Substring(0, i) + '0' + twos.Substring(i + 1,twos.Length-( i+1)); } else { twos = twos.Substring(0, i) + '1' + twos.Substring(i + 1,twos.Length-(i+1)); odmor; } } // Če ni prekinitve: vsi so 1 kot v 111 ali 11111; // v takem primeru dodajte dodatno 1 na začetku if (i == -1) { twos = '1' + twos; } Console.WriteLine(komplement '1: ' + enice);; Console.WriteLine('2's komplement: ' + dvojke); } // Koda gonilnika public static void Main(String[] args) { String bin = '1100'; printOneAndTwosComplement(bin); } } // To kodo je prispeval 29AjayKumar>

Javascript

> > // Javascript program to print 1's and 2's complement of> // a binary number> > // Returns '0' for '1' and '1' for '0'> function> flip (c) {> return> (c ==> '0'> )?> '1'> :> '0'> ;}> > // Print 1's and 2's complement of binary number> // represented by 'bin'> function> printOneAndTwosComplement(bin)> {> > var> n = bin.length;> > var> i;> > > var> ones, twos;> > ones = twos => ''> ;> > > // for ones complement flip every bit> > for> (i = 0; i ones += flip(bin[i]); // for two's complement go from right to left in // ones complement and if we get 1 make, we make // them 0 and keep going left when we get first // 0, make that 1 and go out of loop twos = ones; twos = twos.split('') for (i = n - 1; i>= 0; i--) { if (ones[i] == '1') twos[i] = '0'; else { twos[i] = '1'; odmor; } } twos = twos.join('') // Če ni prekinitve: vsi so 1 kot v 111 ali 11111; // v takem primeru dodajte dodatno 1 na začetku if (i == -1) twos = '1' + twos; document.write( '1's komplement: ' + enice + ' '); document.write( '2's komplement: ' + dvojke + ' '); } // Program gonilnika var bin = '1100'; printOneAndTwosComplement(bin);>

Izhod:

1's complement: 0011 2's complement: 0100 

Časovna zapletenost: O(n)

Pomožni prostor: O(1)