#practiceLinkDiv {weergave: geen! belangrijk; }Gegeven drie cijfers een b En M waar 1<=bm<=10^6 En 'A' kan erg groot zijn en maximaal bevatten 10^6 cijfers. De taak is om te vinden (a^b)%m .
Voorbeelden:
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 Zoek (a^b)%m Probeer het!
Dit probleem is in principe gebaseerd op modulaire rekenkunde. Wij kunnen schrijven (a^b) % m als (a%m) * (a%m) * (a%m) * ... (a%m) b keer . Hieronder staat een algoritme om dit probleem op te lossen:
topologieën
- Omdat 'a' erg groot is, lees 'a' dus als string.
- Nu proberen we 'a' te verminderen. We nemen modulo van 'a' door m één keer, dat wil zeggen; ans = a % m nu op deze manier ans=a%m ligt tussen geheel getalbereik 1 tot 10^6, dat wil zeggen; 1<= a%m <= 10^6.
- Vermenigvuldig nu jaar door b-1 keer en neem tegelijkertijd het modificatieresultaat van de tussenliggende vermenigvuldiging met m omdat de tussenliggende vermenigvuldiging van jaren kan het bereik van een geheel getal overschrijden en zal een verkeerd antwoord opleveren.
// C++ program to find (a^b) mod m for a large 'a' #include
// 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
# 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(1b): ans = (ans*mul) % m return ans # Driver program to run the case a = '987584345091051645734583954832576' b m = 3 11 print (ApowBmodM(a b m))
// 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 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. ?> <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>
Uitvoer
10
Tijdcomplexiteit: O(alleen(a)+b)
Hulpruimte: O(1)
Efficiënte aanpak: Bovenstaande vermenigvuldigingen kunnen worden herleid tot log b door te gebruiken snelle modulaire machtsverheffing waarbij we het resultaat berekenen door de binaire representatie van de exponent (B) . Als het ingestelde bit is 1 we vermenigvuldigen de huidige waarde van de basis met het resultaat en kwadrateren de waarde van de basis voor elke recursieve aanroep.
Recursieve code:
C++14// C++ program to find (a^b) mod m for a large 'a' with an // efficient approach. #include
// 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 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# 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
<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>
Uitvoer
10
Tijdcomplexiteit: O(len(a)+ log b)
Hulpruimte: O(logboek)
Ruimte-efficiënte iteratieve code:
tel verschillende sqlC++14
// C++ program to find (a^b) mod m for a large 'a' #include
// 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 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# 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 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
Uitvoer
10
Tijdcomplexiteit: O(len(a)+ log b)
Hulpruimte: O(1)
Geval: Wanneer zowel 'a' als 'b' erg groot zijn.
We kunnen ook dezelfde aanpak implementeren als beide 'A' En 'B' was erg groot. In dat geval hadden we eerst genomen tegen ervan met M met behulp van onze aModM functie. Geef het dan door aan bovenstaande ApowBmodM recursieve of iteratieve functie om het gewenste resultaat te verkrijgen.
Recursieve code:
Java-constanteC++14
#include
/*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 numint 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 along blong 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(am); long remainderB = MOD(bm); System.out.println(ModExponent(remainderA remainderB m)); } } // This code is contributed by aadityapburujwale
# 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# 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 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
Uitvoer
546081867
Tijdcomplexiteit: O(len(a)+len(b)+log b)
Hulpruimte: O(log b)
Ruimte-efficiënte iteratieve code:
C++14// C++ program to find (a^b) mod m for a large 'a' #include
/*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 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 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# 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
Uitvoer
546081867
Tijdcomplexiteit: O(len(a)+len(b)+ log b)
Hulpruimte: O(1)
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