Trouvez (a^b)%m où 'a' est très grand
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#practiceLinkDiv { display : aucun !important; } Étant donné trois nombres un b et m où 1 <=bm <=10^6 et 'un' peut être très volumineux et contenir jusqu'à 10^6 chiffres. La tâche est de trouver (a^b)%m .
Exemples :
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 Trouver (a^b)%m Essayez-le !
Ce problème est essentiellement basé sur l’arithmétique modulaire. Nous pouvons écrire (a^b) %m comme (a%m) * (a%m) * (a%m) * ... (a%m) b fois . Ci-dessous un algorithme pour résoudre ce problème :
- Puisque « a » est très grand, lisez donc « a » sous forme de chaîne.
- Maintenant, nous essayons de réduire « a ». Nous prenons le modulo de « a » par m une fois, c'est-à-dire ; ans = a % m de cette façon maintenant ans=a%m se situe entre une plage entière de 1 à 10 ^ 6, c'est-à-dire ; 1 <= a%m <= 10^6.
- Maintenant, multipliez ans par b-1 fois et prendre simultanément le mod du résultat de multiplication intermédiaire avec m car la multiplication intermédiaire de ans peut dépasser la plage des nombres entiers et cela produira une mauvaise réponse.
// 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>
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10
Complexité temporelle : O(seulement(a)+b)
Espace auxiliaire : O(1)
Approche efficace : Les multiplications ci-dessus peuvent être réduites à journal b en utilisant exponentiation modulaire rapide où nous calculons le résultat par la représentation binaire de l'exposant (b) . Si le bit défini est 1 nous multiplions la valeur actuelle de la base par le résultat et mettons au carré la valeur de la base pour chaque appel récursif.
Code récursif :
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>
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10
Complexité temporelle : O(len(a)+ log b)
Espace auxiliaire : O (logb)
Code itératif économe en espace :
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
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10
Complexité temporelle : O(len(a)+ log b)
Espace auxiliaire : O(1)
Cas : lorsque « a » et « b » sont tous deux très grands.
Nous pouvons également mettre en œuvre la même approche si les deux 'un' et 'b' était très grand. Dans ce cas, nous aurions d'abord pris contre de cela avec m en utilisant notre aModM fonction. Ensuite, transmettez-le à ce qui précède ApowBmodM fonction récursive ou itérative pour obtenir le résultat requis.
Code récursif :
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
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546081867
Complexité temporelle : O(len(a)+len(b)+log b)
Espace auxiliaire : O (logb)
Code itératif économe en espace :
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
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546081867
Complexité temporelle : O(len(a)+len(b)+log b)
Espace auxiliaire : O(1)
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