Int h(int n) { if (n == 0) return 0; else return h(n/10) + n%10;} What...
Int h(int n)
{
if (n == 0)
return 0;
else
return h(n/10) + n%10;}
What is the output for each function call
(a) Function call h(736). Output:
(b) Function call h(12345). Output:
Answer options:
|
16,15 |
|
|
26,25 |
|
|
36,35 |
|
|
46,45 |
Homework Answers
Add Answer to:
Int h(int n)
{
if (n == 0)
return 0;
else
return h(n/10) + n%10;}
What...
---> JAVA PROGRAM Public Static int EXX (int n,int x){ If (x==0){ return 1; } else...
---> JAVA PROGRAM Public Static int EXX (int n,int x){ If (x==0){ return 1; } else If (n==1) { return 1; } else { return (n* EXX(n,x-1)); * what is the output when (5,0) *what is the output when (8,1) what is the output when (4,3)
1. public int function(int x, int n) { if (n == 0) return 1; return x...
1. public int function(int x, int n) { if (n == 0) return 1; return x * function(x, n -1); } function(3,3) - What is the expected output? 3 12 9 27 2. int fun(int x) { if(x == 0) return 1; else return fun(x - 1); } fun(4) 18 1 24 4 3. Which one of the following calls results 6? int mystery(int n){ if (n == 1) return 1; else return n * mystery(n - 1); } mystery(3)...
#include <stdio.h> int josephus(int n, int k) { if (n == 1) return 1; else...
#include <stdio.h> int josephus(int n, int k) { if (n == 1) return 1; else /* The position returned by josephus(n - 1, k) is adjusted because the recursive call josephus(n - 1, k) considers the original position k%n + 1 as position 1 */ return (josephus(n - 1, k) + k-1) % n + 1; } // Driver Program to test above function int main() { int n = 14; int k = 2; printf("The chosen place...
How to prove G（n）=n+1 in this algorithm？ 1. if (n 0) 2. return 1 3. else if (n1) f 4. return 2 5. else if (n 2) 6. return 3 7. else if (n3) t 8. return 4 else f 9. int OGnew int[n 11 10. G[O]1 1...
How to prove G（n）=n+1 in this algorithm？
1. if (n 0) 2. return 1 3. else if (n1) f 4. return 2 5. else if (n 2) 6. return 3 7. else if (n3) t 8. return 4 else f 9. int OGnew int[n 11 10. G[O]1 12. G[2]3 13. G[3]4 14. int i:-4 15. while (i<n) t 16. if (i mod 20) else ( 20. return G[n]
1. if (n 0) 2. return 1 3. else if (n1) f...
b) Consider the following code. public static int f(int n) if (n == 1) return 0;...
b) Consider the following code. public static int f(int n) if (n == 1) return 0; else if (n % 2 == 0). return g(n/2); else return g(n+1); public static int g(int n) int r = n % 3; if (r == 0) return f(n/3); else if (r == 1) return f(n+2); else return f(2 * n); // (HERE) public static void main(String[] args) { int x = 3; System.out.println(f(x)); (1) (5 points) Draw the call stack as it would...
What does this function return if it is called with n = 4? int f(int n)...
What does this function return if it is called with n = 4? int f(int n) { if (n == 0) return 1; else return f(n-1) + 1; } Can you show the steps as well please?
Consider the following recursive definition of a factorial function. int factorial ( int n) { ...
Consider the following recursive definition of a factorial function. int factorial ( int n) { if ( n == 0 || n ==1 ) return 1; else return n * factorial (n-1); } Suppose the function factorial (4) is invoke. Trace through the function call, explicitly show how the factorial function is repeatedly called and what is the value of n in each call. Also show the value returned by each call. Give an...
Given these methods: METHOD math1: public int math1( int n ) { if (n <= 1) { return 1; } // if...
Given these methods: METHOD math1: public int math1( int n ) { if (n <= 1) { return 1; } // if else { return ( n * 2 ) + math1( n-1 ); } // else } // math1 METHOD math2: public int math2( int n ) { if (n <= 1) { return 1; } // if else { return n + math1( n ) * math2( n/2 ); } // else } // math2 (a) Set up...
What is the base case? -------------------------------------------------- int fibo(int n) { if(n <= 2){ return...
What is the base case? -------------------------------------------------- int fibo(int n) { if(n <= 2){ return n-1; } else{ return fibo(n-1) + fibo(n-2); } } --------------------------------------------------------------------------- int getFibo(int n){ if(n<=2){ return n-1; } else{ int term[n+1], i; term[0] = 0, term[1] = 1; for(i = 2;i= term[i] = term[i-1]+term[i-2]; } return term[n]; } }
int is_prime(int n) { if (n <= 1) return 0; if (n % 2 == 0...
int is_prime(int n) { if (n <= 1) return 0; if (n % 2 == 0 && n > 2) return 0; for (int i = 3; i < n / 2; i += 2) if (n % i == 0) return 0; return 1; } What do the three lines in this function mean?
How to prove G（n）=n+1 in this algorithm？
1. if (n 0) 2. return 1 3. else if (n1) f 4. return 2 5. else if (n 2) 6. return 3 7. else if (n3) t 8. return 4 else f 9. int OGnew int[n 11 10. G[O]1 12. G[2]3 13. G[3]4 14. int i:-4 15. while (i<n) t 16. if (i mod 20) else ( 20. return G[n]
1. if (n 0) 2. return 1 3. else if (n1) f...
b) Consider the following code. public static int f(int n) if (n == 1) return 0; else if (n % 2 == 0). return g(n/2); else return g(n+1); public static int g(int n) int r = n % 3; if (r == 0) return f(n/3); else if (r == 1) return f(n+2); else return f(2 * n); // (HERE) public static void main(String[] args) { int x = 3; System.out.println(f(x)); (1) (5 points) Draw the call stack as it would...
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