#practiceLinkDiv { display: none !important; }Doti trīs skaitļi a b un m kur 1<=bm<=10^6 un "a" var būt ļoti liels un satur līdz 10^6 cipariem. Uzdevums ir atrast (a^b)%m .
Piemēri:
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 Atrast (a^b)%m Izmēģiniet to!
Šī problēma pamatā ir balstīta uz moduļu aritmētiku. Mēs varam rakstīt (a^b) % m kā (a%m) * (a%m) * (a%m) * ... (a%m) b reizes . Tālāk ir sniegts algoritms šīs problēmas risināšanai:
- Tā kā “a” ir ļoti liels, lasiet “a” kā virkni.
- Tagad mēs cenšamies samazināt “a”. Mēs ņemam modulo 'a' ar m vienreiz, t.i.; ans = a % m šādā veidā tagad ans=a%m atrodas starp veselu skaitļu diapazonu no 1 līdz 10^6, t.i.; 1<= a%m <= 10^6.
- Tagad reiziniet gadiem autors b-1 reizes un vienlaikus ņem mod no starpreizināšanas rezultātu ar m, jo starpposma reizināšanu gadiem var pārsniegt vesela skaitļa diapazonu, un tas radīs nepareizu atbildi.
// 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>
Izvade
10
Laika sarežģītība: O(tikai(a)+b)
java anonīmā funkcija
Palīgtelpa: O(1)
Efektīva pieeja: Iepriekš minētos reizinājumus var samazināt līdz žurnāls b izmantojot ātra modulāra kāpināšana kur mēs aprēķinām rezultātu ar eksponenta bināro attēlojumu (b) . Ja iestatītais bits ir 1 mēs reizinām pašreizējo bāzes vērtību līdz rezultātam un katra rekursīvā izsaukuma bāzes vērtību kvadrātā.
Rekursīvais kods:
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>
Izvade
10
Laika sarežģītība: O(len(a)+ log b)
kad iznāca win 7
Palīgtelpa: O(logb)
Kosmosa efektīvas iteratīvais kods:
C++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
Izvade
10
Laika sarežģītība: O(len(a)+ log b)
Palīgtelpa: O(1)
Gadījums: ja gan “a”, gan “b” ir ļoti lieli.
Mēs varam arī īstenot to pašu pieeju, ja abi "a" un "b" bija ļoti liels. Tādā gadījumā mēs vispirms būtu paņēmuši pret no tā ar m izmantojot mūsu aModM funkciju. Pēc tam nododiet to iepriekšminētajam ApowBmodM rekursīva vai iteratīva funkcija, lai iegūtu vajadzīgo rezultātu.
Rekursīvais kods:
tādas vietnes kā coomeetC++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
Izvade
546081867
Laika sarežģītība: O(len(a)+len(b)+log b)
virkne konvertēt uz int java
Palīgtelpa: O(logb)
Kosmosa efektīvas iteratīvais kods:
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
Izvade
546081867
Laika sarežģītība: O(len(a)+len(b)+ log b)
Palīgtelpa: O(1)
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