Funkcija pow() se uporablja za izračun moči danega celega števila. V tem članku bomo s pomočjo programa razumeli, kako izračunati moč celega števila brez uporabe funkcije pow() v C.

Uporaba zanke for za določanje moči danega celega števila

Predstavljajte si, da morate poiskati a ^ b. Najlažja metoda je množenje a z b-krat z uporabo zanke.

• Naj bo a ^ b vhod. Osnova je a, medtem ko je eksponent b.
• Začnite s potenco 1.
• Z uporabo zanke b-krat izvedite naslednja navodila
• moč = moč * a
• Elektroenergetski sistem ima končno rešitev, a ^ b.

Razumejmo zgornji pristop bolje s primerom programa v C:

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 # 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 ( &apos; enter the number and its exponential :  n  n &apos; ) ; scanf ( &apos; % d % d &apos; , &amp; n , &amp; exp ) ; exp1 = exp ; // storing original value for future use // same as while ( ( - - exp ) ! = - 1 ) while ( exp - - &gt; 0 ) { value * = n ; // multiply n to itself exp times } printf ( &apos;  n  n % d ^ % d = % l l d  n  n &apos; , 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 &apos;a&apos; 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&apos;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 ( &apos; enter Base : &apos; ) ; scanf ( &apos; % d &apos; , &amp; base ) ; printf ( &apos; enter Power : &apos; ) ; scanf ( &apos; % d &apos; , &amp; exponent ) ; printf ( &apos; % d ^ % d = % d &apos; , 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) &amp; O(N) space complexity, internal stack.</p> <hr></=>

Pojasnilo

Zgornja koda ima časovno kompleksnost O (N), kjer je N eksponent. O je kompleksnost prostora (1).

Uporaba zanke Medtem:

 # 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; } 

Izhod:

 enter the number and its exponential : 5 4 5 ^ 6 = 625 .......................... Process executed in 0.11 seconds Press any key to continue. 

Pojasnilo

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Long Long Int je dvakrat večji od Long Int. Specifikator formata za long long int je percent lld.

Uporaba rekurzije za iskanje moči danega celega števila

Predpostavimo, da je a ^ b vhod. Moč 'a' se bo z vsakim rekurzivnim klicem povečala za eno. Da dobimo a ^ b, dvakrat pokličemo rekurzivno funkcijo b.

• Naj bo Pow ( a, b ) rekurzivna funkcija, uporabljena za izračun a ^ b.
• Preprosto vrni 1, če je b == 0; sicer vrni Pow (a, b -1) * a.

Razumejmo zgornji pristop bolje s primerom programa v 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; } 

Izhod:

 Enter Base: 5 Enter Power: 4 5 ^ 4 = 625 .......................... Process executed in 1.22 seconds Press any key to continue. 

Pojasnilo:

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V zgornjem primeru kode v C bi bila časovna kompleksnost eksponent N, O(N) in O(N) kompleksnost prostora, notranji sklad.