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:
string.format java niz
# 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></=> 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
vrstni red po naključnem sql
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:
mvc java
V zgornjem primeru kode v C bi bila časovna kompleksnost eksponent N, O(N) in O(N) kompleksnost prostora, notranji sklad.