Trouver la racine cubique d'un nombre
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#practiceLinkDiv { display : aucun !important; } Étant donné un nombre n, trouvez la racine cubique de n.
Exemples :
Input: n = 3 Output: Cubic Root is 1.442250 Input: n = 8 Output: Cubic Root is 2.000000
Nous pouvons utiliser recherche binaire . Nous définissons d’abord l’erreur e. Disons 0,0000001 dans notre cas. Les principales étapes de notre algorithme de calcul de la racine cubique d'un nombre n sont :
- Initialiser début = 0 et fin = n
- Calculer mi = (début + fin)/2
- Vérifiez si la valeur absolue de (n - mid*mid*mid) < e. If this condition holds true then mid is our answer so return mid.
- Si (mid*mid*mid)>n alors définissez end=mid
- Si (milieu*milieu*milieu)
Vous trouverez ci-dessous la mise en œuvre de l’idée ci-dessus.
// C++ program to find cubic root of a number // using Binary Search #include using namespace std ; // Returns the absolute value of n-mid*mid*mid double diff ( double n double mid ) { if ( n > ( mid * mid * mid )) return ( n - ( mid * mid * mid )); else return (( mid * mid * mid ) - n ); } // Returns cube root of a no n double cubicRoot ( double n ) { // Set start and end for binary search double start = 0 end = n ; // Set precision double e = 0.0000001 ; while ( true ) { double mid = ( start + end ) / 2 ; double error = diff ( n mid ); // If error is less than e then mid is // our answer so return mid if ( error <= e ) return mid ; // If mid*mid*mid is greater than n set // end = mid if (( mid * mid * mid ) > n ) end = mid ; // If mid*mid*mid is less than n set // start = mid else start = mid ; } } // Driver code int main () { double n = 3 ; printf ( 'Cubic root of %lf is %lf n ' n cubicRoot ( n )); return 0 ; }
Java // Java program to find cubic root of a number // using Binary Search import java.io.* ; class GFG { // Returns the absolute value of n-mid*mid*mid static double diff ( double n double mid ) { if ( n > ( mid * mid * mid )) return ( n - ( mid * mid * mid )); else return (( mid * mid * mid ) - n ); } // Returns cube root of a no n static double cubicRoot ( double n ) { // Set start and end for binary search double start = 0 end = n ; // Set precision double e = 0.0000001 ; while ( true ) { double mid = ( start + end ) / 2 ; double error = diff ( n mid ); // If error is less than e then mid is // our answer so return mid if ( error <= e ) return mid ; // If mid*mid*mid is greater than n set // end = mid if (( mid * mid * mid ) > n ) end = mid ; // If mid*mid*mid is less than n set // start = mid else start = mid ; } } // Driver program to test above function public static void main ( String [] args ) { double n = 3 ; System . out . println ( 'Cube root of ' + n + ' is ' + cubicRoot ( n )); } } // This code is contributed by Pramod Kumar
Python3 # Python 3 program to find cubic root # of a number using Binary Search # Returns the absolute value of # n-mid*mid*mid def diff ( n mid ) : if ( n > ( mid * mid * mid )) : return ( n - ( mid * mid * mid )) else : return (( mid * mid * mid ) - n ) # Returns cube root of a no n def cubicRoot ( n ) : # Set start and end for binary # search start = 0 end = n # Set precision e = 0.0000001 while ( True ) : mid = ( start + end ) / 2 error = diff ( n mid ) # If error is less than e # then mid is our answer # so return mid if ( error <= e ) : return mid # If mid*mid*mid is greater # than n set end = mid if (( mid * mid * mid ) > n ) : end = mid # If mid*mid*mid is less # than n set start = mid else : start = mid # Driver code n = 3 print ( 'Cubic root of' n 'is' round ( cubicRoot ( n ) 6 )) # This code is contributed by Nikita Tiwari.
C# // C# program to find cubic root // of a number using Binary Search using System ; class GFG { // Returns the absolute value // of n - mid * mid * mid static double diff ( double n double mid ) { if ( n > ( mid * mid * mid )) return ( n - ( mid * mid * mid )); else return (( mid * mid * mid ) - n ); } // Returns cube root of a no. n static double cubicRoot ( double n ) { // Set start and end for // binary search double start = 0 end = n ; // Set precision double e = 0.0000001 ; while ( true ) { double mid = ( start + end ) / 2 ; double error = diff ( n mid ); // If error is less than e then // mid is our answer so return mid if ( error <= e ) return mid ; // If mid * mid * mid is greater // than n set end = mid if (( mid * mid * mid ) > n ) end = mid ; // If mid*mid*mid is less than // n set start = mid else start = mid ; } } // Driver Code public static void Main () { double n = 3 ; Console . Write ( 'Cube root of ' + n + ' is ' + cubicRoot ( n )); } } // This code is contributed by nitin mittal.
PHP // PHP program to find cubic root // of a number using Binary Search // Returns the absolute value // of n - mid * mid * mid function diff ( $n $mid ) { if ( $n > ( $mid * $mid * $mid )) return ( $n - ( $mid * $mid * $mid )); else return (( $mid * $mid * $mid ) - $n ); } // Returns cube root of a no n function cubicRoot ( $n ) { // Set start and end // for binary search $start = 0 ; $end = $n ; // Set precision $e = 0.0000001 ; while ( true ) { $mid = (( $start + $end ) / 2 ); $error = diff ( $n $mid ); // If error is less // than e then mid is // our answer so return mid if ( $error <= $e ) return $mid ; // If mid*mid*mid is // greater than n set // end = mid if (( $mid * $mid * $mid ) > $n ) $end = $mid ; // If mid*mid*mid is // less than n set // start = mid else $start = $mid ; } } // Driver Code $n = 3 ; echo ( 'Cubic root of $n is ' ); echo ( cubicRoot ( $n )); // This code is contributed by nitin mittal. ?>
JavaScript < script > // Javascript program to find cubic root of a number // using Binary Search // Returns the absolute value of n-mid*mid*mid function diff ( n mid ) { if ( n > ( mid * mid * mid )) return ( n - ( mid * mid * mid )); else return (( mid * mid * mid ) - n ); } // Returns cube root of a no n function cubicRoot ( n ) { // Set start and end for binary search let start = 0 end = n ; // Set precision let e = 0.0000001 ; while ( true ) { let mid = ( start + end ) / 2 ; let error = diff ( n mid ); // If error is less than e then mid is // our answer so return mid if ( error <= e ) return mid ; // If mid*mid*mid is greater than n set // end = mid if (( mid * mid * mid ) > n ) end = mid ; // If mid*mid*mid is less than n set // start = mid else start = mid ; } } // Driver Code let n = 3 ; document . write ( 'Cube root of ' + n + ' is ' + cubicRoot ( n )); < /script>
Sortir:
Cubic root of 3.000000 is 1.442250
Complexité temporelle : O (JOURNAL)
Espace auxiliaire : O(1)
