Codeforces Round #467 (Div. 2)

C. Save Energy!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after kminutes after turning on.

During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.

It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.

Input

The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).

Output

Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.

Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if .

Examples

input

Copy

3 2 6

output

6.5

input

Copy

4 2 20

output

20.0

Note

In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for . Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for . Thus, after four minutes the chicken will be cooked for . Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready .

In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.

题意大概是说 煮东西需要T分钟 但炉子会在开启k分钟后关闭 小明同学每d分钟就会进房间查看炉子有没有关 如果炉子关闭了 那他就会重启开启 不然就不会做什么 炉子关闭后 煮东西的效率会减少一半

这道题主要是找到周期

当d > k 时 周期为d 一个周期能煮的时间是k + (d - k) / 2.0;

当d <= k时

 当k是d的倍数的时候 直接输出T就行

不然周期就是 最小的 大于k的 d的倍数 假定它是t  一个周期能煮的时间是 (t - k)/2 + k

当求完周期后，就可以算出T中有多少个周期，剩下不够一个周期的可以推算出要多少时间

贴上Java 代码

import java.util.Scanner;

public class Main {

	public static void main(String args[]) {

		Scanner input = new Scanner(System.in);

		double k = input.nextDouble();
		double d = input.nextDouble();
		double T = input.nextDouble();
		double t;

		double temp;

		if (d > k) {
			temp = k + (d - k) / 2.0;
			t = d;
		} else {
			if (k % d == 0) {
				System.out.println(T);
				input.close();
				return;
			} else {
				t = (long) (k / d + 1) * d;
				temp = k + (t - k) / 2.0;
			}
		}

		
		if (T % t == 0) {
			System.out.println(T / temp * t);
		} else {

			double sum = (long) (T / temp) * t;
			T = T - (long) (T / temp) * temp;

			if (T > k)
				sum += k + (T - k) * 2.0;
			else
				sum += T;
			System.out.println(sum);
		}

		input.close();
	}

}