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最新推荐文章于 2019-08-04 21:15:35 发布

xiaoxiede_wo

最新推荐文章于 2019-08-04 21:15:35 发布

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分类专栏：思路题

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本文介绍了一个烹饪鸡的时间计算问题，考虑到炉子会在特定时间内自动关闭并由Julia定期重启。文章提供了一种解决该问题的有效算法，并通过两个例子展示了如何计算烹饪所需的确切时间。

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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 $\text{[math]}$ .

Examples
inputCopy
3 2 6
output
6.5
inputCopy
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 $\text{[math]}$ . Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for $\text{[math]}$ . Thus, after four minutes the chicken will be cooked for $\text{[math]}$ . Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready $\text{[math]}$ .

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.

本来的思路写法因为精度问题，有的样例过不了，看了博客思路很腻害

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll k,d,t,T;
    while(~scanf("%lld%lld%lld",&k,&d,&t))
    {
        T=k+(k%d==0? 0:(k/d+1)*d-k);//周期长度
        double T1=k+(T-k)*0.5;//根据成熟度求得的周期
        ll num=t/T1;//需要num个周期能熟。
        double ans=num*T;//小于等于t的周期的最大倍数
        double st=t-num*T1;//还不熟部分
        //处理不熟部分
        if(st<k) ans+=st;
        else ans+=(k+(st-k)*2);
        printf("%.1lf\n",ans);
    }
}