Answer:
Explanation;
else
System.out.println("f1 and f2 are not equal");
switch (input.charAt(0)
{
case '+':
f3 = f1.add(f2);
System.out.println("f1+f2=" + f3);
break;
case '-':
f3 = f1.subtract(f2);
System.out.println("f1-f2=" + f3);
break;
case '*':
f3 = f1.multiply(f2);
System.out.println("f1*f2="+f3);
break;
case '/':
f3 = f1.divide(f2);
System.out.println("f1/f2="+f3);
break;
default:
System.out.println("Illegal command: " + input );
break;
}
}// end of while loop
} // end of main
}
Note ; this is the last part of the programme check the attachment from 1-5 this is the 6th .
Can you help me with homework and the answer is a circle
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
Answer:
import java.util.*;
public class MyClass {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.print("Input a word: ");
String userinput = input.nextLine();
for(int i =0;i<userinput.length();i+=2) {
System.out.print(userinput.charAt(i));
}
}
}
Explanation:
This line prompts user for input
System.out.print("Input a word: ");
This declares a string variable named userinput and also gets input from the user
String userinput = input.nextLine();
The following iterates through every other character of userinput from the first using iteration variable i and i is incremented by 2
for(int i =0;i<userinput.length();i+=2) {
This prints characters at i-th position
System.out.print(userinput.charAt(i));
Answer:
Answer is B: Early silent film producers used many of the same acting techniques popularized in vaudeville performances, even using popular vaudeville stars as actors in their productions.
Explanation: