Funktionen pow() bruges til at beregne potensen af et givet heltal. Nu i denne artikel vil vi ved hjælp af et program forstå, hvordan man beregner styrken af et heltal uden at bruge funktionen pow() i C.
Brug for loop til at bestemme styrken af et givet heltal
Forestil dig, at du skal finde en ^ b. Den nemmeste metode er at gange a med b gange ved hjælp af en løkke.
- Lad a ^ b være input. Grundlaget er a, mens eksponenten er b.
- Start med en potens på 1.
- Brug en løkke til at udføre følgende instruktioner b gange
- power = power * a
- Elsystemet har den endelige løsning, a ^ b.
Lad os forstå ovenstående tilgang bedre med et eksempel på et program i C:
# include # include # include # include # include int Pow ( int a , int b ) { int power = 1 , i ; for ( i = 1 ; i <= b ; + i ) { power="power" * a } return int main ( long base , exponent printf ' enter : scanf % d & ^ pow < pre> <p> <strong>Output:</strong> </p> <pre> Enter Base: 5 Enter Power: 3 5 ^ 3 = 125 .......................... Process executed in 3.22 seconds Press any key to continue. </pre> <p> <strong>Explanation</strong> </p> <p>The code above has an O (N) time complexity, where N is the exponent. O is the space complexity (1).</p> <h3>Using While loop:</h3> <pre> # include # include # include # include # include int main ( ) { int n , exp , exp1 ; long long int value = 1 ; printf ( ' enter the number and its exponential : n n ' ) ; scanf ( ' % d % d ' , & n , & exp ) ; exp1 = exp ; // storing original value for future use // same as while ( ( - - exp ) ! = - 1 ) while ( exp - - > 0 ) { value * = n ; // multiply n to itself exp times } printf ( ' n n % d ^ % d = % l l d n n ' , n , exp1 , value ) ; return 0; } </pre> <p> <strong>Output:</strong> </p> <pre> enter the number and its exponential : 5 4 5 ^ 6 = 625 .......................... Process executed in 0.11 seconds Press any key to continue. </pre> <p> <strong>Explanation</strong> </p> <p>Long Long Int is twice as large as Long Int. The format specifier for long long int is percent lld.</p> <h2>Using Recursion to find the Power of Given Integer</h2> <p>Assume that a ^ b is the input. The power of 'a' will increase by one with each recursive call. To obtain a ^ b, we call the recursive function b twice.</p> <ul> <li>Let Pow ( a, b ) be the recursive function used to calculate a ^ b.</li> <li>Simply return 1 if b == 0; else, return Pow (a, b -1) * a.</li> </ul> <p> <strong>Let's understand the above approach better with an example of a program in C:</strong> </p> <pre> # include # include # include # include # include int Pow ( int a , int b ) { if ( b = = 0 ) return 1 ; else return Pow ( a , b - 1 ) * X ; } int main ( ) { long long int base , exponent ; printf ( ' enter Base : ' ) ; scanf ( ' % d ' , & base ) ; printf ( ' enter Power : ' ) ; scanf ( ' % d ' , & exponent ) ; printf ( ' % d ^ % d = % d ' , base , exponent , Pow ( base , exponent ) ) ; return 0; } </pre> <p> <strong>Output:</strong> </p> <pre> Enter Base: 5 Enter Power: 4 5 ^ 4 = 625 .......................... Process executed in 1.22 seconds Press any key to continue. </pre> <p> <strong>Explanation:</strong> </p> <p>In the above example of a code in C, time complexity would be exponent N, O(N) & O(N) space complexity, internal stack.</p> <hr></=> Forklaring
Ovenstående kode har en O (N) tidskompleksitet, hvor N er eksponenten. O er rumkompleksiteten (1).
Brug af While loop:
# include # include # include # include # include int main ( ) { int n , exp , exp1 ; long long int value = 1 ; printf ( ' enter the number and its exponential : n n ' ) ; scanf ( ' % d % d ' , & n , & exp ) ; exp1 = exp ; // storing original value for future use // same as while ( ( - - exp ) ! = - 1 ) while ( exp - - > 0 ) { value * = n ; // multiply n to itself exp times } printf ( ' n n % d ^ % d = % l l d n n ' , n , exp1 , value ) ; return 0; } Produktion:
enter the number and its exponential : 5 4 5 ^ 6 = 625 .......................... Process executed in 0.11 seconds Press any key to continue.
Forklaring
Long Long Int er dobbelt så stor som Long Int. Formatspecifikationen for long long int er procent lld.
Brug af rekursion til at finde kraften i et givet heltal
Antag, at a ^ b er input. Effekten af 'a' vil stige med én for hvert rekursivt kald. For at opnå a ^ b kalder vi den rekursive funktion b to gange.
- Lad Pow ( a, b ) være den rekursive funktion, der bruges til at beregne a ^ b.
- Du skal blot returnere 1, hvis b == 0; ellers returner Pow (a, b -1) * a.
Lad os forstå ovenstående tilgang bedre med et eksempel på et program i C:
# include # include # include # include # include int Pow ( int a , int b ) { if ( b = = 0 ) return 1 ; else return Pow ( a , b - 1 ) * X ; } int main ( ) { long long int base , exponent ; printf ( ' enter Base : ' ) ; scanf ( ' % d ' , & base ) ; printf ( ' enter Power : ' ) ; scanf ( ' % d ' , & exponent ) ; printf ( ' % d ^ % d = % d ' , base , exponent , Pow ( base , exponent ) ) ; return 0; } Produktion:
Enter Base: 5 Enter Power: 4 5 ^ 4 = 625 .......................... Process executed in 1.22 seconds Press any key to continue.
Forklaring:
I ovenstående eksempel på en kode i C ville tidskompleksitet være eksponent N, O(N) & O(N) rumkompleksitet, intern stak.