Answer:
The ouput of the given code is :
22
is "Tom's age.
Explanation:
Here in this code the variable user_name and user_age are initialized to "Tom" and 22 respectively as statement is given in the question i.e cout << user_age << " \nis " + user_name << "'s age.";.This line will print the user_age i.e 22 after that the control moves to the next line and print is "Tom's age.
Following are the code in c++
#include <iostream> // header file
#include <string>
using namespace std;
int main() // main function
{
string user_name="Tom";
int user_age= 22;
cout << user_age << " \nis " + user_name << "'s age.";
return 0;
}
Output:
22
is "Tom's age.
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.
A looping is a set of instructions which is repeated a certain number of times until a condition is met. hlo dai k xa halkhabar
