Answer:
To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.
Similarly check for 400th year and multiple 0f 4. The following C program describes the function.
#include<stdio.h>
#include<stdbool.h>
bool is_leap_year(int year);
void main()
{
int y;
bool b;
printf("Enter the year in yyyy format: e.g. 1999 \n");
scanf("%d", &y); // taking the input year in yyyy format.
b= is_leap_year(y); //calling the function and returning the output to b
if(b==true)
{
printf("Thae given year is a leap year \n");
}
else
{
printf("The given year is not a leap year \n");
}
}
bool is_leap_year(int year)
{
if(year%100==0) //every 100th year
{
if(year%400==0) //every 400th year
{
return true;
}
else
{
return false;
}
}
if(year%4==0) //is a multiple of 4
{
return true;
}
else
{
return false;
}
}
Explanation:
Output is given as image
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:
int main() {
int _2dArray[32][32];
for (int i = 0; i < 32; i++) {
for (int j = 0; j < 32; j++) {
_2dArray[i][j] = j + i * 32;
}
}
return 0;
}
Explanation:
Here is a generic C/C++ 2d array traversal and main function example. The rest you'll have to figure out based on what kind of app you're making.
Good luck!
Using a linear search to find a value that is stored in the last element of an array of 20,000 elements, 20,000 element(s) must be compared.
