Codeforces 937C--Save Energy!

xiaoxiao2021-03-01  5

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 k minutes 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:这个人在用烤炉烤一只鸡,这个鸡需要t秒在烤炉里,或者是烤炉处在保温状态2*t秒.这个人会在每d秒的时候来看一下烤炉,如果烤炉保温了,然后这只鸡没做好的话就按一下烤炉,让烤炉烤起来烤炉运作一次会持续k秒。

题解:

我们假设鸡需要t秒烤的热量或者是2t秒保温的热量

我们看完这个题后我们应该很容易会想到这样一个过程:

    一开始的时候 人按下按钮后就走了,然后过了一段时间后,当烤炉刚好保温的时候 人过来了,把烤炉按上,接着烤。

那么问题来了,这样的循环有几个?我们应该怎么样求这个循环所花的时间和提供给鸡的热量。

在这个循环里 烤炉烤了几秒,保温了几秒?

我们设两个变量t1 t2分别表示烤炉烤的时间和保温的时间

显而易见的 t1=k,t2=d-(k%d);

这里提供一个疑问? 当k能被d整除的时候

这种情况请大家先思考几分钟,

如果能整除的话 t2=0

然后我们看到提示里 这个式子,我们可以思考思考,可不可以把这个等式变一下,让这个等式好求解。

我们可以把题目给的变量给统一

    鸡需要t秒烤的热量--->t*=2;

    烤炉烤的时候提供的热量--->k*2

    烤炉保温的时候提供的热量--->t2

一次循环需要的时间就是t1+t2

一次循环提供的热量就是k*2+t2

如果有剩余的需要提供的热量的话

分类讨论一下就行了,这里就不具体细讲

分为一下两种情况

(1) 烤炉还在运行的时候 鸡就烤好了

(2)烤炉在保温的时候 鸡才好

这个数据有点大,超过了int的范围了,所以我们应该用long long来定义变量。

最后附上我的AC代码

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; long long k,d,t; // k 烤炉的时间 // d 人走的时间间隔 // t 所需时间 int main(){ scanf("%I64d%I64d%I64d",&k,&d,&t); t*=2; long long t1=k,t2=d-(k%d); if(t2==d) t2=0; long long xun=2*k+t2; double te=(t/xun); double ans=(1.0*te)*(t1+t2); t%=xun; //printf("%lld %lld %lld %lf",t1,t2,xun,ans); if(t<=2*k){ ans+=t/2.0; } else{ t-=2*k; ans+=k; ans+=t; } printf("%.1lf\n",ans); return 0; }
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