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  • Codeforces Round #471 (Div. 2)

                                A. Feed the cat

    After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.

    At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.

    Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.

    Input

    The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.

    The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).

    Output

    Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.

    Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

    Examples
    Input
    Copy
    19 00
    255 1 100 1
    Output
    25200.0000
    Input
    Copy
    17 41
    1000 6 15 11
    Output
    1365.0000
    Note

    In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.

    In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.

    注意是不是要多买一个buns。

     1 #pragma warning(disable:4996)
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<iostream>
     5 #include<algorithm>
     6 using namespace std;
     7 typedef long long ll;
     8 
     9 int hh, mm, H, D, C, N;
    10 
    11 int main()
    12 {
    13     double x, y;
    14     while (cin >> hh >> mm >> H >> D >> C >> N) {
    15         if (H % N) x = (H / N + 1)*C;
    16         else x = (H / N)*C;
    17         if (hh >= 20) printf("%.4lf
    ", x*0.8);
    18         else {
    19             int temp = H + ((20 - hh) * 60 - mm)*D;
    20             if (temp % N) y = (temp / N + 1)*C*0.8;
    21             else y = (temp / N)*C*0.8;
    22             printf("%.4lf
    ", min(x, y));
    23         }
    24     }
    25     return 0;
    26 }
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  • 原文地址:https://www.cnblogs.com/zgglj-com/p/8637906.html
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