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:
numbers = 1:1:100;
for num=numbers
remainder3 = rem(num,3);
remainder5 = rem(num,5);
if remainder3==0
disp("Yee")
else
if remainder3 == 0 && remainder5 == 0
disp ("Yee-Haw")
else
if remainder5==0
disp("Haw")
else
disp("Not a multiple of 5 or 4")
end
end
end
end
Explanation:
- Initialize the numbers variable from 1 to 100.
- Loop through the all the numbers and find their remainders.
- Check if a number is multiple of 5, 3 or both and display the message accordingly.
Answer:
Application
Explanation:
Application software comes in many forms like apps, and even on computers. The software is most common on computers of all kinds while mobile applications are most common on cellular devices.
Answer:
1. Date 2. It will appear to the right of the selected column.
