Alarm Clock | Editorial by manikasharma

import java.util.Scanner;
        public class Main {
        	public static void main(String[] args) {
        		Scanner sc = new Scanner(System.in);
        		int t = sc.nextInt();
        		while (t-- > 0) {
        			int a = sc.nextInt();
        			int b = sc.nextInt();
        			int c = sc.nextInt();
        			int d = sc.nextInt();
        			if (b >= a)
        				System.out.println(b);
        			else {
        				if (d >= c)
        					System.out.println(-1);
        				else
        					System.out.println(((a - b + c - d - 1) / (c - d)) * (long) c + b);
        			}
        		}
        	}
        }
        
 In this problem we can see that if the value of b (which is the time before the first alarm goes off) is greater than the value of a which is the required amount tof sleep polycarp needs we can say that he got the required amount of sleep and print b.
 And if the the value of d is greater than c which means the time after which the alarm is set again is less than the time after which CArp sleeps again then we can comprehend that polycarp will not get the desired amount of sleep.
 So print -1 for that case.
 If none of this happens then this means that Finally, if Polycarp resets his alarm x times then he ends up with b+x⋅(c−d) minutes of sleep in total and ends up spending b+x⋅c minutes of time. We know that b+x⋅(c−d) should be greater or equal to a and x

should be the smallest possible.

b+x⋅(c−d)≥a↔x⋅(c−d)≥a−b↔x≥a−bc−d

Thus, the smallest possible integer x
is equal to ⌈a−bc−d⌉. And the answer is b+x⋅c.

Alarm Clock | Editorial by manikasharma

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