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Question 2.An array containing the given n-element integers and the sum of the consecutive elements in the aray with the integer k Find how many of them are divisible without remainder to the given integer k. • The array contains n elements ( array = [array [(0], array (1..., array (n-1]] ). •k is a positive integer • Sum of consecutive elements for any ( first , next ) pair array (first] + array (next] if o first

Question 2.An array containing the given n-element integers and the sum of the consecutive elements in the aray with the integer k Find how many of them are divisible without remainder to the given integer k. • The array contains n elements ( array = [array [(0], array (1..., array (n-1]] ). •k is a positive integer • Sum of consecutive elements for any ( first , next ) pair array (first] + array (next] if o first

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question 2.An array containing the given n-element integers and the sum of the consecutive elements in the aray with the integer k Find how many of them are divisible without remainder to the given integer k. • The array contains n elements ( array = [array [(0], array (1..., array (n-1]] ). •k is a positive integer • Sum of consecutive elements for any ( first , next ) pair array (first] + array (next] if o first <next Write the Java code that returns the result for the integer array (array) and integer (k) entered in this context. V Define the section to be caleulated as the Java object (" Question2 "). V Question2 ; Take the note string as input. o public Question2 ( int [] input ) V Take the value of " k" as input in the " countSumPairs " method to calculate and how many retum the sum of the pairs is divided by the value "k" without any remainder. o public int countSumPairs ( int k ) Code sample: public class Question2 { public Question2 ( int [] input ) { I/ Complete the countSumPairs function below public int countSumPairs ( int k ) ( For example: Let us assume that an array of positive integers looks like this: If Numbers = {1, 3, 2, 6, 1, 2} and k = 3 , the output will be given on the screen. • (0, 2) → array (0] + array [2] = 1 + 2 = 3 • (0, 5) → array [0] + array [5] = 1 + 2 = 3 •(1, 3) → array [1] + array [3] = 3 + 6 = 9 •(2, 4) → array [2] + array [4] = 2 + 1 = 3 • (4, 5) → array [4] + array [5] = 1 + 2 = 3 A total of 5 element pairs were found for the value k = 3.
Transcribed Image Text:Question 2.An array containing the given n-element integers and the sum of the consecutive elements in the aray with the integer k Find how many of them are divisible without remainder to the given integer k. • The array contains n elements ( array = [array [(0], array (1..., array (n-1]] ). •k is a positive integer • Sum of consecutive elements for any ( first , next ) pair array (first] + array (next] if o first <next Write the Java code that returns the result for the integer array (array) and integer (k) entered in this context. V Define the section to be caleulated as the Java object (" Question2 "). V Question2 ; Take the note string as input. o public Question2 ( int [] input ) V Take the value of " k" as input in the " countSumPairs " method to calculate and how many retum the sum of the pairs is divided by the value "k" without any remainder. o public int countSumPairs ( int k ) Code sample: public class Question2 { public Question2 ( int [] input ) { I/ Complete the countSumPairs function below public int countSumPairs ( int k ) ( For example: Let us assume that an array of positive integers looks like this: If Numbers = {1, 3, 2, 6, 1, 2} and k = 3 , the output will be given on the screen. • (0, 2) → array (0] + array [2] = 1 + 2 = 3 • (0, 5) → array [0] + array [5] = 1 + 2 = 3 •(1, 3) → array [1] + array [3] = 3 + 6 = 9 •(2, 4) → array [2] + array [4] = 2 + 1 = 3 • (4, 5) → array [4] + array [5] = 1 + 2 = 3 A total of 5 element pairs were found for the value k = 3.
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