Covid Run Codechef October long challenge problem solution| Codechef October long challenge problem solution

Covid Run Codechef October long  challenge problem solution| Codechef October long  challenge problem solution -

This is very first and easy problem of Codechef October long  challenge problem solution| Codechef October long  challenge problem solution.

Problem Statement-

Covid-19$19$ is spreading fast! There are N$N$ cities, numbered from 0$0$ to (N−1)$(N-1)$, arranged in a circular manner. City 0$0$ is connected to city 1$1$, 1$1$ to 2$2$, …$\dots$, city (N−2)$(N-2)$ to city (N−1)$(N-1)$, and city (N−1)$(N-1)$ to city 0$0$.

The virus is currently at city X$X$. Each day, it jumps from its current city, to the city K$K$ to its right, i.e., from city X$X$ to the city (X+K)%N$(X+K)\mathrm{\%}N$. As the virus jumps, the cities in between don't get infected. Cities once infected stay infected. You live in city Y$Y$. Find if it will reach your city eventually. If it will, print YES, else print NO.

Input:

• The first line of the input consists of an integer T$T$, the number of test cases.
• The first and only line of each test case contains four space-separated integers - N$N$, K$K$, X$X$ and Y$Y$, denoting the number of cities, the size of jumps, Covid's current city, and the city that you live in, respectively.

Output:

For each test case, in a new line, print YES if Covid shall reach your city after a finite number of days, else print NO.

Constraints

• 1≤T≤100$1\le T\le 100$
• 1≤N≤1000$1\le N\le 1000$
• 0≤X,Y≤N−1$0\le X,Y\le N-1$
• 0≤K≤1000$0\le K\le 1000$

Subtasks

• Subtask 1 - 100% - Original constraints

Sample Input:

2
6 2 5 3
12 3 4 2

Sample Output:

YES
NO

Explanation:

• In the first sample, Covid starts at city 5$5$, then goes to city 1$1$, and then from city 1$1$ to city 3$3$. Thus, it reaches the city that you live in.

• In the second sample, Covid starts at city 4$4$, goes to city 7$7$, then 10$10$, 1$1$, then 4$4$, 7$7$, 10$10$, …$\dots$, and so on. It never reaches city 2$2$.

Problem link->https://www.codechef.com/OCT20B/problems/CVDRUN

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Let's directly jump into solution 

according to problem statement take input t= total number of test cases .

next line contain n,k,xand y

take input for n,k,x and y.

virus is currently at city x.

spreading rate of virus is k means virus can jump k city from its currenty position x.

Question says that you are living currently at y city so you have to tell that virus may infect you or not.

Logic-

Approach of this question is very simple simply you have to jump k step from current city and if this goes out of (N-1) city  it will 

return back to city (n+k)%n th city.

dont try to think too much follow steps as mentioned in question it is called bruete force approach.

My approach-

My approach says that if the virus currently at city x in the next step it will jump to (x+k)%n th city.

What i will do i will maintain a  flag variable that will tell me that virus can reach to place y or not.

step 1-If virus already at position y. so i have to simply return true.otherwise, I will move form current city to x+k th city.in this case i will mark my flag as true and terminate the loop.

step 2- if virus reach again at x position it mean it will never reach to city y and the person who is sitting in city y is safe.

in this case i will mark my flag as false and terminate the loop.

finally, i will print the result according to the flag

            if(z){

                cout<<"YES"<<endl;

            }

            else{

                cout<<"NO"<<endl;

            }

Entire code solution is given below-