C - Save Energy!

Love The Way You Lie 2021-11-09 04:02 158阅读 0赞

题目链接：[https://vjudge.net/contest/317253\#problem/C][https_vjudge.net_contest_317253_problem_C]  
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

3 2 6  
Output  
6.5  
Input  
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.

分析:  
如果d < k，我们可以找到最早的再次需要开灯的时间d。把新d作为d考虑。  
然后，我们考虑d时间内，对鸡产生的效果sum是多少。  
这里为了避免小数计算，我们统一把开火的时候产生的效果记为2(每分钟)，  
关火记为1（每分钟）。t在开始的时候就可以乘2。  
d分钟内产生的效果sum = 2 \* k + (d - k);  
我们在考虑需要产生t效果值，会有多少个sum。  
即时间 time = t / sum \* d;  
可是这样还可能会剩一些热量没有加热。  
所以需要对剩下的热量再考虑  
mod = (t % sum);  
这个时候，我们考虑mod是在k时间内就可以完成，还是需要在d时间内完成。这二者是不同的。

if(mod <= 2 * k)
            time += mod / 2.0;
        else
        {
            time += k;
            ll mul = mod - 2 * k;
            time += mul;
        }

code

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
    using namespace std;
    typedef long long ll;
    
    ll k,d,t;
    
    int main()
    {
        scanf("%lld%lld%lld",&k,&d,&t);
        if(k > d)
        {
            ll v = k / d;
            if(k % d)
                v ++;
            d = d * v;
        }
        t = 2 * t;
        ll a = k;
        ll b = d - k;
        ll sum = a * 2 + b;///d 分钟内产生的效果
    
        double time = t / sum * d;
    
        ll mod = t % sum; ///还差mod能量
    
        if(mod <= 2 * k)
            time += mod / 2.0;
        else
        {
            time += k;
            ll mul = mod - 2 * k;
            time += mul;
        }
        printf("%.1lf\n",time);
    }

[https_vjudge.net_contest_317253_problem_C]: https://vjudge.net/contest/317253#problem/C