CodeForces_937C Save Energy!(貪心)

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.

http://codeforces.com/contest/937/problem/C

題目：

輸入三個數字k,d,t,分別表明微波爐自動關的時間、Julia回來的間隔（若是關了就從新打開）、進程須要時間，若是在微波爐開的時候加熱效率就是1:1，關的時候效率是1:2,問你最少須要多少妙才能完成加熱。

解題報告：

這題是以能覆蓋k的d的倍數爲週期（這裏值得複習），而後分開再處理餘數部分就好了下面是代碼

#include<cstdio>
typedef long long ll;
#define eps 1e-10
double k,d,t;
int main()
{
    while(scanf("%lf%lf%lf",&k,&d,&t)!=EOF)
	{
		double ti=(long long)((k-1)/d)+1;//恰好覆蓋的d的倍數 
		double p=ti*d;//一個週期的長度 
	    double ans=0;
		double tim=(long long)(2*t/(p+k));//p+k爲了對應2*t 
	    ans+=tim*p;
		double rest=2*t-(p+k)*tim;
	    if (rest<2*k) ans+=rest/2;
		else ans+=k,ans+=rest-2*k;
	    printf("%.1lf",ans);
	}
}