Answer and Explanation:
Using javascript:
function dayof_theweek(){
var TodayDate = window. prompt("enter today's date in the format 'year, month, day' ");
var Datenow=new date(TodayDate);
var Dayofweek=Datenow.getday();
var Days=["monday","Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"];
if(Dayofweek==indexOf(Days[Dayofweek]))
{
document.createTextnode(Days[Dayofweek]);
}
}
The program above uses a date object which uses the method getday to get the day of the week(get day returns an integer from 0 to 6).we then use a comparison operator == to test the condition that returned value Dayoftheweek is same with the index of the array Days and then print to a html document. The program may need improvements such as the fact that errors may arise when proper input isn't given, and therefore must be handled.
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
Answer: The difference between call by value and call by reference is that in call by value the actual parameters are passed into the function as arguments whereas in call by reference the address of the variables are sent as parameters.
Explanation:
Some examples are:
call by value
#include <stdio.h>
void swap(int, int);
int main()
{ int a = 10, b= 20;
swap(a, b);
printf("a: %d, b: %d\n", a, b);
}
void swap(int c, int d)
{
int t;
t = c; c = d; d = t;
}
OUTPUT
a: 10, b: 20
The value of a and b remain unchanged as the values are local
//call by reference
#include <stdio.h>
void swap(int*, int*);
int main()
{
int a = 10, b = 20;
swap(&a, &b); //passing the address
printf("a: %d, b: %d\n", a, b);
}
void swap(int *c, int *d)
{
int t;
t = *c; *c = *d; *d = t;
}
OUTPUT
a: 20, b: 10
due to dereferencing by the pointer the value can be changed which is call by reference
