Raskite (a^b)%m, kur „a“ yra labai didelis
Išbandykite „GfG Practice“. 
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#practiceLinkDiv { display: none !important; } Duoti trys skaičiai a b ir m kur 1 <=bm <=10^6 ir 'a' gali būti labai didelis ir jame gali būti iki 10^6 skaitmenys. Užduotis – surasti (a^b)%m .
Pavyzdžiai:
Input : a = 3 b = 2 m = 4 Output : 1 Explanation : (3^2)%4 = 9%4 = 1 Input : a = 987584345091051645734583954832576 b = 3 m = 11 Output: 10Recommended Practice Rasti (a^b)%m Išbandykite!
Ši problema iš esmės pagrįsta moduline aritmetika. Galime rašyti (a^b) % m kaip (a%m) * (a%m) * (a%m) * ... (a%m) b kartus . Žemiau pateikiamas algoritmas, kaip išspręsti šią problemą:
- Kadangi „a“ yra labai didelis, skaitykite „a“ kaip eilutę.
- Dabar bandome sumažinti „a“. Mes paimame modulo 'a' iš m vieną kartą, t.y.; ans = a % m šiuo būdu dabar ans=a%m yra tarp sveikųjų skaičių diapazono nuo 1 iki 10^6, t.y.; 1 <= a%m <= 10^6.
- Dabar padauginkite metų pateikė b-1 kartų ir vienu metu imti tarpinio daugybos rezultato modą su m, nes tarpinis daugybos rezultatas metų gali viršyti sveikojo skaičiaus diapazoną ir pateiks neteisingą atsakymą.
// C++ program to find (a^b) mod m for a large 'a' #include using namespace std ; // utility function to calculate a%m unsigned int aModM ( string s unsigned int mod ) { unsigned int number = 0 ; for ( unsigned int i = 0 ; i < s . length (); i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s [ i ] - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m unsigned int ApowBmodM ( string & a unsigned int b unsigned int m ) { // Find a%m unsigned int ans = aModM ( a m ); unsigned int mul = ans ; // now multiply ans by b-1 times and take // mod with m for ( unsigned int i = 1 ; i < b ; i ++ ) ans = ( ans * mul ) % m ; return ans ; } // Driver program to run the case int main () { string a = '987584345091051645734583954832576' ; unsigned int b = 3 m = 11 ; cout < < ApowBmodM ( a b m ); return 0 ; }
Java // Java program to find (a^b) mod m for a large 'a' public class GFG { // utility function to calculate a%m static int aModM ( String s int mod ) { int number = 0 ; for ( int i = 0 ; i < s . length (); i ++ ) { // (s[i]-'0') gives the digit // value and form the number number = ( number * 10 ); int x = Character . getNumericValue ( s . charAt ( i )); number = number + x ; number %= mod ; } return number ; } // Returns find (a^b) % m static int ApowBmodM ( String a int b int m ) { // Find a%m int ans = aModM ( a m ); int mul = ans ; // now multiply ans by b-1 times // and take mod with m for ( int i = 1 ; i < b ; i ++ ) ans = ( ans * mul ) % m ; return ans ; } // Driver code public static void main ( String args [] ) { String a = '987584345091051645734583954832576' ; int b = 3 m = 11 ; System . out . println ( ApowBmodM ( a b m )); } } // This code is contributed by Sam007
Python3 # Python program to find (a^b) mod m for a large 'a' def aModM ( s mod ): number = 0 # convert string s[i] to integer which gives # the digit value and form the number for i in range ( len ( s )): number = ( number * 10 + int ( s [ i ])) number = number % m return number # Returns find (a^b) % m def ApowBmodM ( a b m ): # Find a%m ans = aModM ( a m ) mul = ans # now multiply ans by b-1 times and take # mod with m for i in range ( 1 b ): ans = ( ans * mul ) % m return ans # Driver program to run the case a = '987584345091051645734583954832576' b m = 3 11 print ( ApowBmodM ( a b m ))
C# // C# program to find (a^b) mod m // for a large 'a' using System ; class GFG { // utility function to calculate a%m static int aModM ( string s int mod ) { int number = 0 ; for ( int i = 0 ; i < s . Length ; i ++ ) { // (s[i]-'0') gives the digit // value and form the number number = ( number * 10 ); int x = ( int )( s [ i ] - '0' ); number = number + x ; number %= mod ; } return number ; } // Returns find (a^b) % m static int ApowBmodM ( string a int b int m ) { // Find a%m int ans = aModM ( a m ); int mul = ans ; // now multiply ans by b-1 times // and take mod with m for ( int i = 1 ; i < b ; i ++ ) ans = ( ans * mul ) % m ; return ans ; } // Driver Code public static void Main () { string a = '987584345091051645734583954832576' ; int b = 3 m = 11 ; Console . Write ( ApowBmodM ( a b m )); } } // This code is contributed by Sam007
PHP // PHP program to find (a^b) // mod m for a large 'a' // utility function to // calculate a%m function aModM ( $s $mod ) { $number = 0 ; for ( $i = 0 ; $i < strlen ( $s ); $i ++ ) { // (s[i]-'0') gives the digit // value and form the number $number = ( $number * 10 + ( $s [ $i ] - '0' )); $number %= $mod ; } return $number ; } // Returns find (a^b) % m function ApowBmodM ( $a $b $m ) { // Find a%m $ans = aModM ( $a $m ); $mul = $ans ; // now multiply ans by // b-1 times and take // mod with m for ( $i = 1 ; $i < $b ; $i ++ ) $ans = ( $ans * $mul ) % $m ; return $ans ; } // Driver code $a = '987584345091051645734583954832576' ; $b = 3 ; $m = 11 ; echo ApowBmodM ( $a $b $m ); return 0 ; // This code is contributed by nitin mittal. ?>
JavaScript < script > // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function aModM ( s mod ) { let number = 0 ; for ( let i = 0 ; i < s . length ; i ++ ) { // (s[i]-'0') gives the digit // value and form the number number = ( number * 10 ); let x = ( s [ i ] - '0' ); number = number + x ; number %= mod ; } return number ; } // Returns find (a^b) % m function ApowBmodM ( a b m ) { // Find a%m let ans = aModM ( a m ); let mul = ans ; // Now multiply ans by b-1 times // and take mod with m for ( let i = 1 ; i < b ; i ++ ) ans = ( ans * mul ) % m ; return ans ; } // Driver Code let a = '987584345091051645734583954832576' ; let b = 3 m = 11 ; document . write ( ApowBmodM ( a b m )); // This code is contributed by souravghosh0416 < /script>
Išvestis
10
Laiko sudėtingumas: O (tik (a) + b)
Pagalbinė erdvė: O(1)
Efektyvus požiūris: Aukščiau pateiktus daugybas galima sumažinti iki žurnalas b naudojant greitas modulinis eksponentas kur rezultatą apskaičiuojame dvejetainiu eksponento vaizdu b) . Jei nustatytas bitas yra 1 Mes padauginame dabartinę bazės vertę iki rezultato ir kiekvieno rekursinio skambučio bazės vertę padauginame kvadratu.
Rekursyvus kodas:
C++14 // C++ program to find (a^b) mod m for a large 'a' with an // efficient approach. #include using namespace std ; typedef long long ll ; // Reduce the number B to a small number // using Fermat Little ll MOD ( string num int mod ) { ll res = 0 ; for ( int i = 0 ; i < num . length (); i ++ ) res = ( res * 10 + num [ i ] - '0' ) % mod ; return res ; } ll ModExponent ( ll a ll b ll m ) { ll result ; if ( a == 0 ) return 0 ; else if ( b == 0 ) return 1 ; else if ( b & 1 ) { result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result * result ) % m + m ) % m ; } return ( result % m + m ) % m ; } int main () { // String input as b is very large string a = '987584345091051645734583954832576' ; // String input as b is very large ll b = 3 ; ll m = 11 ; ll remainderA = MOD ( a m ); cout < < ModExponent ( remainderA b m ); return 0 ; }
Java // Java program to find (a^b) mod m for a large 'a' with an // efficient approach. public class GFG { // Reduce the number B to a small number // using Fermat Little static long MOD ( String num long mod ) { long res = 0 ; for ( int i = 0 ; i < num . length (); i ++ ) { res = ( res * 10 + num . charAt ( i ) - '0' ) % mod ; } return res ; } // Calculate the ModExponent of the given large number // 'a' static long ModExponent ( long a long b long m ) { long result ; if ( a == 0 ) { return 0 ; } else if ( b == 0 ) { return 1 ; } else if ( b % 2 != 0 ) { result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result * result ) % m + m ) % m ; } return ( result % m + m ) % m ; } public static void main ( String [] args ) { // String input as a is very large String a = '987584345091051645734583954832576' ; long b = 3 ; long m = 11 ; long remainderA = MOD ( a m ); System . out . println ( ModExponent ( remainderA b m )); } } // The code is contributed by Gautam goel (gautamgoel962)
Python3 # Python3 program to find (a^b) mod m # for a large 'a' # Utility function to calculate a%m def MOD ( s mod ): res = 0 for i in range ( len ( s )): res = ( res * 10 + int ( s [ i ])) % mod return res # Returns find (a^b) % m def ModExponent ( a b m ): if ( a == 0 ): return 0 elif ( b == 0 ): return 1 elif ( b % 2 != 0 ): result = a % m result = result * ModExponent ( a b - 1 m ) else : result = ModExponent ( a b / 2 m ) result = (( result * result ) % m + m ) % m return ( result % m + m ) % m # Driver Code a = '987584345091051645734583954832576' b = 3 m = 11 remainderA = MOD ( a m ) print ( ModExponent ( remainderA b m )) # This code is contributed by phasing17
C# // C# program to find (a^b) mod m for a large 'a' with an // efficient approach. using System ; using System.Collections.Generic ; public class GFG { // Reduce the number B to a small number // using Fermat Little static long MOD ( string num long mod ) { long res = 0 ; for ( int i = 0 ; i < num . Length ; i ++ ) { res = ( res * 10 + num [ i ] - '0' ) % mod ; } return res ; } // Calculate the ModExponent of the given large number // 'a' static long ModExponent ( long a long b long m ) { long result ; if ( a == 0 ) { return 0 ; } else if ( b == 0 ) { return 1 ; } else if ( b % 2 != 0 ) { result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result * result ) % m + m ) % m ; } return ( result % m + m ) % m ; } // Driver Code public static void Main ( string [] args ) { // String input as a is very large string a = '987584345091051645734583954832576' ; long b = 3 ; long m = 11 ; // Function Call long remainderA = MOD ( a m ); Console . WriteLine ( ModExponent ( remainderA b m )); } } // The code is contributed by phasing17
JavaScript < script > // JavaScript program to find (a^b) mod m // for a large 'a' // Utility function to calculate a%m function MOD ( s mod ) { var res = 0 ; for ( var i = 0 ; i < s . length ; i ++ ) { res = ( res * 10 + ( s [ i ] - '0' )) % mod ; } return res ; } // Returns find (a^b) % m function ModExponent ( a b m ) { var result ; if ( a == 0 ) { return 0 ; } else if ( b == 0 ) { return 1 ; } else if ( b % 2 != 0 ) { result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result * result ) % m + m ) % m ; } return ( result % m + m ) % m ; } // Driver Code let a = '987584345091051645734583954832576' ; let b = 3 m = 11 ; var remainderA = MOD ( a m ); document . write ( ModExponent ( remainderA b m )); // This code is contributed by shinjanpatra. < /script>
Išvestis
10
Laiko sudėtingumas: O(len(a)+ log b)
Pagalbinė erdvė: O(logb)
Kosmosui efektyvus kartotinis kodas:
C++14 // C++ program to find (a^b) mod m for a large 'a' #include using namespace std ; typedef long long ll ; // utility function to calculate a%m and b%m ll aModM ( string s ll mod ) { ll number = 0 ; for ( ll i = 0 ; i < s . length (); i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s [ i ] - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m ll ApowBmodM ( ll x ll y ll m ) { ll res = 1 ; while ( y ) { if ( y & 1 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x * x ) % m + m ) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case int main () { string a = '987584345091051645734583954832576' ; ll b = 3 ; ll m = 11 ; // Find a%m ll x = aModM ( a m ); cout < < ApowBmodM ( x b m ); return 0 ; }
Java // Java program to find (a^b) mod m for a large 'a' import java.util.* ; class GFG { // utility function to calculate a%m and b%m static long aModM ( String s long mod ) { long number = 0 ; for ( int i = 0 ; i < s . length (); i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s . charAt ( i ) - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m static long ApowBmodM ( long x long y long m ) { long res = 1 ; while ( y > 0 ) { if (( y & 1 ) != 0 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x * x ) % m + m ) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case public static void main ( String [] args ) { String a = '987584345091051645734583954832576' ; long b = 3 ; long m = 11 ; // Find a%m long x = aModM ( a m ); System . out . println ( ApowBmodM ( x b m )); } } // This code is contributed by phasing17
Python3 # Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%m def aModM ( s mod ): number = 0 ; for i in range ( len ( s )): # int(s[i]) gives the digit value and form # the number number = ( number * 10 + int ( s [ i ])); number %= mod ; return number ; # Returns find (a^b) % m def ApowBmodM ( x y m ): res = 1 ; while ( y > 0 ): if ( y & 1 ): res = ( res * x ) % m ; y = y >> 1 ; x = (( x * x ) % m + m ) % m ; return ( res % m + m ) % m ; # Driver program to run the case a = '987584345091051645734583954832576' ; b = 3 ; m = 11 ; # Find a%m x = aModM ( a m ); print ( ApowBmodM ( x b m )); # This code is contributed by phasing17
C# // C# program to find (a^b) mod m for a large 'a' using System ; class GFG { // utility function to calculate a%m and b%m static long aModM ( string s long mod ) { long number = 0 ; for ( int i = 0 ; i < s . Length ; i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s [ i ] - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m static long ApowBmodM ( long x long y long m ) { long res = 1 ; while ( y > 0 ) { if (( y & 1 ) != 0 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x * x ) % m + m ) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case public static void Main ( string [] args ) { string a = '987584345091051645734583954832576' ; long b = 3 ; long m = 11 ; // Find a%m long x = aModM ( a m ); Console . WriteLine ( ApowBmodM ( x b m )); } } // This code is contributed by phasing17
JavaScript // JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%m function aModM ( s mod ) { let number = 0 ; for ( var i = 0 ; i < s . length ; i ++ ) { // parseInt(s[i]) gives the digit value and form // the number number = ( number * 10 + parseInt ( s [ i ])); number %= mod ; } return number ; } // Returns find (a^b) % m function ApowBmodM ( x y m ) { let res = 1 ; while ( y ) { if ( y & 1 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x * x ) % m + m ) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case let a = '987584345091051645734583954832576' ; let b = 3 ; let m = 11 ; // Find a%m let x = aModM ( a m ); console . log ( ApowBmodM ( x b m )); // This code is contributed by phasing17
Išvestis
10
Laiko sudėtingumas: O(len(a)+ log b)
Pagalbinė erdvė: O(1)
Atvejis: kai „a“ ir „b“ yra labai dideli.
Taip pat galime įgyvendinti tą patį metodą, jei abu 'a' ir "b" buvo labai didelis. Tokiu atveju pirmiausia būtume paėmę prieš iš jo su m naudojant mūsų aModM funkcija. Tada perduokite jį aukščiau ApowBmodM rekursinė arba pasikartojanti funkcija norint gauti reikiamą rezultatą.
Rekursyvus kodas:
C++14 #include using namespace std ; typedef long long ll ; // Reduce the number B to a small number // using Fermat Little ll MOD ( string num int mod ) { ll res = 0 ; for ( int i = 0 ; i < num . length (); i ++ ) res = ( res * 10 + num [ i ] - '0' ) % mod ; return res ; } ll ModExponent ( ll a ll b ll m ) { ll result ; if ( a == 0 ) return 0 ; else if ( b == 0 ) return 1 ; else if ( b & 1 ) { result = a % m ; result = result * ModExponent ( a b -1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result % m ) * ( result % m )) % m ; } return ( result % m + m ) % m ; } int main () { // String input as b is very large string a = '987584345091051645734583954832576' ; // String input as b is very large string b = '467687655456765756453454365476765' ; ll m = 1000000007 ; ll remainderA = MOD ( a m ); ll remainderB = MOD ( b m ); cout < < ModExponent ( remainderA remainderB m ); return 0 ; }
Java /*package whatever //do not write package name here */ import java.io.* ; class GFG { // Reduce the number B to a small number // using Fermat Little. static long MOD ( String num int mod ) { long res = 0 ; for ( int i = 0 ; i < num . length (); i ++ ) { res = ( res * 10 + num . charAt ( i ) - '0' ) % mod ; } return res ; } static long ModExponent ( long a long b long m ){ long result = 0 ; if ( a == 0 ) return 0 ; else if ( b == 0 ) return 1 ; else if (( b & 1 ) == 1 ){ result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result % m ) * ( result % m )) % m ; } return ( result % m + m ) % m ; } public static void main ( String [] args ) { // String input as b is very large String a = '987584345091051645734583954832576' ; // String input as b is very large String b = '467687655456765756453454365476765' ; int m = 1000000007 ; long remainderA = MOD ( a m ); long remainderB = MOD ( b m ); System . out . println ( ModExponent ( remainderA remainderB m )); } } // This code is contributed by aadityapburujwale
Python3 # Python3 program to implement the approach # Reduce the number B to a small number # using Fermat Little def MOD ( num mod ): res = 0 ; for i in range ( len ( num )): res = ( res * 10 + int ( num [ i ])) % mod ; return res ; def ModExponent ( a b m ): if ( a == 0 ): return 0 ; elif ( b == 0 ): return 1 ; elif ( b & 1 ): result = a % m ; result = result * ModExponent ( a b - 1 m ); else : b = b // 2 result = ModExponent ( a b m ); result = (( result % m ) * ( result % m )) % m ; return ( result % m + m ) % m ; # String input as b is very large a = '987584345091051645734583954832576' ; # String input as b is very large b = '467687655456765756453454365476765' ; m = 1000000007 ; remainderA = ( MOD ( a m )); remainderB = ( MOD ( b m )); print ( ModExponent ( remainderA remainderB m )); # This code is contributed by phasing17
C# // C# program to implement the approach using System ; using System.Collections.Generic ; class GFG { // Reduce the number B to a small number // using Fermat Little. static long MOD ( string num int mod ) { long res = 0 ; for ( int i = 0 ; i < num . Length ; i ++ ) { res = ( res * 10 + num [ i ] - '0' ) % mod ; } return res ; } static long ModExponent ( long a long b long m ) { long result = 0 ; if ( a == 0 ) return 0 ; else if ( b == 0 ) return 1 ; else if (( b & 1 ) == 1 ) { result = a % m ; result = result * ModExponent ( a b - 1 m ); } else { result = ModExponent ( a b / 2 m ); result = (( result % m ) * ( result % m )) % m ; } return ( result % m + m ) % m ; } public static void Main ( string [] args ) { // String input as b is very large string a = '987584345091051645734583954832576' ; // String input as b is very large string b = '467687655456765756453454365476765' ; int m = 1000000007 ; long remainderA = MOD ( a m ); long remainderB = MOD ( b m ); Console . WriteLine ( ModExponent ( remainderA remainderB m )); } } // This code is contributed by phasing17
JavaScript // JavaScript program to implement the approach // Reduce the number B to a small number // using Fermat Little function MOD ( num mod ) { let res = 0 ; for ( var i = 0 ; i < num . length ; i ++ ) res = ( res * 10 + parseInt ( num [ i ])) % mod ; return res ; } function ModExponent ( a b m ) { let result ; if ( a == 0n ) return 0n ; else if ( b == 0n ) return 1n ; else if ( b & 1n ) { result = a % m ; result = result * ModExponent ( a b - 1n m ); } else { b = b / 2n - ( b % 2n ); result = ModExponent ( a b m ); result = (( result % m ) * ( result % m )) % m ; } return ( result % m + m ) % m ; } // String input as b is very large let a = '987584345091051645734583954832576' ; // String input as b is very large let b = '467687655456765756453454365476765' ; let m = 1000000007 ; let remainderA = BigInt ( MOD ( a m )); let remainderB = BigInt ( MOD ( b m )); console . log ( ModExponent ( remainderA remainderB BigInt ( m ))); // This code is contributed by phasing17
Išvestis
546081867
Laiko sudėtingumas: O(len(a)+len(b)+log b)
Pagalbinė erdvė: O(logb)
Kosmosui efektyvus kartotinis kodas:
C++14 // C++ program to find (a^b) mod m for a large 'a' #include using namespace std ; typedef long long ll ; // utility function to calculate a%m and b%m ll aModM ( string s ll mod ) { ll number = 0 ; for ( ll i = 0 ; i < s . length (); i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s [ i ] - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m ll ApowBmodM ( string & a string & b ll m ) { ll res = 1 ; // Find a%m ll x = aModM ( a m ); // Find b%m ll y = aModM ( b m ); while ( y ) { if ( y & 1 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x % m ) * ( x % m )) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case int main () { string a = '987584345091051645734583954832576' ; string b = '467687655456765756453454365476765' ; ll m = 1000000007 ; cout < < ApowBmodM ( a b m ); return 0 ; }
Java /*package whatever //do not write package name here */ import java.io.* ; class GFG { // utility function to calculate a%m and b%m static long aModM ( String s long mod ){ long number = 0 ; for ( int i = 0 ; i < s . length (); i ++ ) { // (s.charAt(i)-'0') gives the digit value and form // the number number = ( number * 10 + ( s . charAt ( i ) - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m static long ApowBmodM ( String a String b long m ) { long res = 1 ; // Find a%m long x = aModM ( a m ); // Find b%m long y = aModM ( b m ); while ( y > 0 ) { if (( y & 1 ) == 1 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x % m ) * ( x % m )) % m ; } return ( res % m + m ) % m ; } public static void main ( String [] args ) { String a = '987584345091051645734583954832576' ; String b = '467687655456765756453454365476765' ; long m = 1000000007 ; System . out . println ( ApowBmodM ( a b m )); } } // This code is contributed by aadityapburujwale
Python3 # Python3 program to find (a^b) mod m for a large 'a' # utility function to calculate a%m and b%m def aModM ( s mod ): number = 0 for i in range ( len ( s )): # (s[i]-'0') gives the digit value and form # the number number = ( number * 10 + ( int ( s [ i ]))) number %= mod return number # Returns find (a^b) % m def ApowBmodM ( a b m ): res = 1 # Find a%m x = aModM ( a m ) # Find b%m y = aModM ( b m ) while ( y > 0 ): if ( y & 1 ): res = ( res * x ) % m y = y >> 1 x = (( x % m ) * ( x % m )) % m return ( res % m + m ) % m # Driver program to run the case a = '987584345091051645734583954832576' b = '467687655456765756453454365476765' m = 1000000007 print ( ApowBmodM ( a b m )) # This code is contributed by phasing17
JavaScript // JavaScript program to find (a^b) mod m for a large 'a' // utility function to calculate a%m and b%m function aModM ( s mod ) { let number = 0n ; for ( let i = 0 ; i < s . length ; i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10n + BigInt ( parseInt ( s [ i ]))); number %= mod ; } return number ; } // Returns find (a^b) % m function ApowBmodM ( a b m ) { let res = 1n ; // Find a%m let x = BigInt ( aModM ( a m )); // Find b%m let y = BigInt ( aModM ( b m )); while ( y > 0n ) { if ( y & 1n ) res = ( res * x ) % m ; y = y >> 1n ; x = (( x % m ) * ( x % m )) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case let a = '987584345091051645734583954832576' ; let b = '467687655456765756453454365476765' ; let m = 1000000007n ; console . log ( ApowBmodM ( a b m )); // This code is contributed by phasing17
C# // C# program to find (a^b) mod m for a large 'a' using System ; using System.Collections.Generic ; class GFG { // utility function to calculate a%m and b%m static long aModM ( string s long mod ) { long number = 0 ; for ( int i = 0 ; i < s . Length ; i ++ ) { // (s[i]-'0') gives the digit value and form // the number number = ( number * 10 + ( s [ i ] - '0' )); number %= mod ; } return number ; } // Returns find (a^b) % m static long ApowBmodM ( string a string b long m ) { long res = 1 ; // Find a%m long x = aModM ( a m ); // Find b%m long y = aModM ( b m ); while ( y != 0 ) { if (( y & 1 ) != 0 ) res = ( res * x ) % m ; y = y >> 1 ; x = (( x % m ) * ( x % m )) % m ; } return ( res % m + m ) % m ; } // Driver program to run the case public static void Main ( string [] args ) { string a = '987584345091051645734583954832576' ; string b = '467687655456765756453454365476765' ; long m = 1000000007 ; Console . WriteLine ( ApowBmodM ( a b m )); } } // This code is contributed by phasing17
Išvestis
546081867
Laiko sudėtingumas: O(len(a)+len(b)+ log b)
Pagalbinė erdvė: O(1)
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