Answer:
import java.util.Scanner;
public class TestClock {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
System.out.print("Enter favorite color:");
String word1 = in.next();
System.out.print("Enter pet's name:");
String word2 = in.next();
System.out.print("Enter a number:");
int num = in.nextInt();
System.out.println("you entered: "+word1+" "+word2+" "+num);
}
}
Explanation:
Using Java Programming language
- Import the Scanner class
- create an object of the scanner class
- Prompt user to enter the values for the variables (word1, word2, num)
- Use String concatenation in System.out.println to display the output as required by the question.
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Answer:
Really depends on your situation. If you specified, I'd have an asnwer
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
