Question

INNER PRODUCT 4E Input Standard input Output Standard output Topic Array & Array Processing! Problem Description The inner product (also known as the dot product or scalar product) of the elements of set A and the elements of set B of size is defined as the sum of the products of corresponding terms. For example, given set A as the array (2, 3, 4, 5, 6, 7, 8, 9; and set B as 6,5, 4, 3, 2, 7, 8, 9), then the inner product of A and B is (2*6) + (3*5) + (4*4) + (5*3)+(6*2)+(7*7) + (8*8) + (9*9) = 264. Input First line of input is an integer N (I <N<100) that represents the number of test case, followed by N test cases. Each test case will start with an integer 1 (1 ST < 20) in a single line that represents the set size. For each of the following two lines there will be T integers representing data in set A and B respectively. Output For each test case, the output contains a line in the format "Case #x: IP", where x is the case number (starting from 1) and IP is the inner product for the test case Sample Input Output Sample Input Sample Output 2 8 2 3 4 5 6 7 8 9 6 5 4 3 2 7 8 9 4 21 41 6 5 83 Case 1: 264 Case #2: 52

Java code

Answer 1

Java Code:

import java.util.*;

class Main {

public static void main(String[] args) {

int t;

Scanner sc=new Scanner(System.in);

t=sc.nextInt();

int a[]=new int[t];

for(int i=0;i<t;i++){

int n;

n=sc.nextInt();

int arr1[]=new int[n];

int arr2[] = new int[n],sum=0;

for(int j=0;j<n;j++)

arr1[j]=sc.nextInt();

for(int j=0;j<n;j++)

arr2[j]=sc.nextInt();

for(int j=0;j<n;j++)

sum+=arr1[j]*arr2[j];

a[i]=sum;

}

for(int i=0;i<t;i++)

System.out.println("Case #"+(i+1)+": "+a[i]);

}

}

E Mai > javac -classpath .-/run_dir/junit-4.12.jar:target/dependency/* n.java java -classpath .:/run_dir/junit-4.12. jar:target/dependency/* Main 2 8 2 3 4 5 6 7 8 9 6 5 4 3 2 7 8 9 4 2 1 4 1 6 583 Case #1: 264 Case #2: 52 int n; Main.java 1 import java.util.*; 2 class Main 3 public static void main(String[] args) { 4 int t; 5 Scanner sc=new Scanner(System.in); 6 t=sc.nextInt(); 7 int a[]=new int[t]; 8 for(int i=0;i<t;i++){ 9 10 n=sc.nextInt(); 11 int arr1[]=new int[n]; 12 int arr2[] = new int[n], sum=8; 13 for(int j=0; j<n; j++) 14 arri[j]=sc.nextInt(); 15 for(int j=0; j<n; j++) 16 arr2[j]=sc.nextInt(); 17 for(int j=0; j<n; j++) 18 sum+=arr1[j]*arr2[j]; 19 a[i]=sum; 20 } 21 for(int i=0;i<t;i++) 22 System.out.println("Case #"+(i+1)+": "+a[i]); 23 24 }

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