Answer:
The answer is A I hope this helps, if not I apologize.
Explanation:
Answer:
import java.util.*;
public class Main {
public static void main(String[] args)
{
Scanner scan = new Scanner();
double budget=0, num=0, total=0;
System.out.println("Your budget for the month? ");
budget=scan.nextDouble();
System.out.println("enter all expense, and after that type -9999 to quit: ");
while(num != -9999)
{
total+=num;
num=scan.nextDouble();
}
if(total<=budget)
{
System.out.println("under budget by ");
System.out.println(budget-total);
}
else
{
System.out.println("over budget by ");
System.out.println(total-budget);
}
}
}
Explanation:
- Take the budget as an input from user and store it to the budget variable.
- Loop until user has entered all his expenses and keep on adding them to the total variable.
- Check If the total is less than or equal to budget or otherwise, and then print the relevant message accordingly.
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:
They allow you to ask more questions and also your rank goes up or down.
Explanation:
