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C++ Recursion program

$10-20 USD

Terminado
Publicado hace alrededor de 7 años

$10-20 USD

Pagado a la entrega
Write individual C++ programs for each function. Implement all of these functions recursively. No loops allowed. No static Variables allowed. bool isPrefix( char target[] , char bigString[] ); // return true iff target is a prefix of bigString. No library functions allowed. // e.g. "abc" is a prefix of "abc" and "abcde" but not of "ab" or "aabc" or "acb" // Notes: isPrefix(w,w) and isPrefix("", w) are always true but isPrefix(y,"") is false unless y is "". // bigString and target are C-strings meaning null-terminated character arrays. bool equal(char s[], char t[]) ; // return true if C-strings s and t are equal int addOdd(int n); // return the sum of all odd numbers from 1 to n double twoToThe(int n); // return 2 to the power n including when n <= 0. int getMax(int nums[], int size); // return the maximum value in nums. int getIndexOfMax(int nums[], int size); // return the index of a maximum element of nums int findLastOccurence(int nums[], int size, int target); // return index of the last occurrence of target or -1 if target is not found. int findFirstOccurence(int nums[], int size, int target); // return index of the first occurrence of target or -1 if target is not found. void printDigits(unsigned int n); // print the digits of n separated by spaces. ie.g. 3415 --> 3 4 1 5. Assume n > 0 void drawLine(int n, char c); // output n occurrences of c on a line e.g. drawLine(4,'X') will output XXXX. void drawRuler(int n); // output a ruler pattern, e.g. drawRuler(4) will output the pattern below. (Call drawLine.) - -- - --- - -- - ---- - -- - --- - -- - string toBase16(unsigned int n); // return a string with the base 16 representation of n. Assume n>0. // 6-10 lines of code should be enough. string reverse(string s); // return a string with the same characters as s in reverse order, e.g. "abc" --> "cba" bool isPalindrome(int nums[], int size); // return true if the values in nums form a palindrome, e.g. {1,2,2,1} or {63,24,63}. bool isSorted(int nums[], int size); // return true if the values in nums are in ascending (or nondescending) order, e.g. {1,1,2,4} but not {1,1,4,2}. unsigned int intLog(unsigned int number, unsigned int base); // return biggest power of base that's <= number. // e.g. intLog(233, 3) is 4 since 3^4 = 81 but 3^5 is 243. // preconditions: base >= 2, number >= 1. int add(int m, int n); // return sum of m and n using ++ and -- operators ( Don't use + or - .) // Be sure to handle the cases where m and/or n is negative. // Hint: m + n = (m+1) + (n-1) = (m-1) + (n+1) int times(int m, int n); // return product of m and n using repeated addition. ( So don't use * .) // e.g. times(3,2) = 3 + times(3,1) = 3 + 3 + times(3,0) = 3 + 3 + 0 = 6 // If n < 0 then use the identity times(m,n) = -times(m,-n) unsigned int skipFactorial(unsigned int n); // n * (n-2) * (n-4) * .... * {1 or 2} and skipFactorial(0) = 1. void echo(); // The user enters a sequence of nonzero integers followed by 0. // The function then outputs the same integers in the original order, not including the 0. // Example: 43 23 88 6 55 0 --> 43 23 88 6 55 // No arrays, strings or other data structures allowed. void echoReverse(); // The user enters a sequence of nonzero integers followed by 0. // The function then outputs the same integers in the reverse order, not including the 0. // Example: 43 23 88 6 55 0 --> 55 6 88 23 43 // No arrays, strings or other data structures allowed.
ID del proyecto: 12749617

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4 propuestas
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Activo hace 7 años

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This looks nice and easy, I know exactly how to write each and every one of these functions already. Just let me at it and it will be done in no time :)
$15 USD en 1 día
5,0 (3 comentarios)
2,9
2,9
4 freelancers están ofertando un promedio de $28 USD por este trabajo
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Hello, I have more that 7 years of experience in Java and .Net. I can do this project. Let me know more about this job. Best regards.
$20 USD en 1 día
5,0 (2 comentarios)
0,0
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Bandera de UNITED STATES
San Jose, United States
5,0
2
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Miembro desde nov 18, 2016

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