CodeForce-734C Anton and Making Potions
 C. Anton and Making Potions
time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Anton
is playing a very interesting computer game, but now he is stuck at one
of the levels. To pass to the next level he has to prepare npotions.

Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.

  1. Spells of this type speed up the preparation time of one potion. There are m spells of this type, the i-th of them costs bi manapoints and changes the preparation time of each potion to ai instead of x.
  2. Spells of this type immediately prepare some number of potions. There are k such spells, the i-th of them costs di manapoints and instantly create ci potions.

Anton can use no more than one spell of the first type and no more than one spell of the second type, and the total number of manapoints spent should not exceed s. Consider that all spells are used instantly and right before Anton starts to prepare potions.

Anton
wants to get to the next level as fast as possible, so he is interested
in the minimum number of time he needs to spent in order to prepare at
least n potions.

Input

The first line of the input contains three integers nmk (1 ≤ n ≤ 2·109, 1 ≤ m, k ≤ 2·105) —
the number of potions, Anton has to make, the number of spells of the
first type and the number of spells of the second type.

The second line of the input contains two integers x and s (2 ≤ x ≤ 2·109, 1 ≤ s ≤ 2·109) — the initial number of seconds required to prepare one potion and the number of manapoints Anton can use.

The third line contains m integers ai (1 ≤ ai < x) — the number of seconds it will take to prepare one potion if the i-th spell of the first type is used.

The fourth line contains m integers bi (1 ≤ bi ≤ 2·109) — the number of manapoints to use the i-th spell of the first type.

There are k integers ci (1 ≤ ci ≤ n) in the fifth line — the number of potions that will be immediately created if the i-th spell of the second type is used. It's guaranteed that ci are not decreasing, i.e. ci ≤ cj if i < j.

The sixth line contains k integers di (1 ≤ di ≤ 2·109) — the number of manapoints required to use the i-th spell of the second type. It's guaranteed that di are not decreasing, i.e. di ≤ dj if i < j.

Output

Print one integer — the minimum time one has to spent in order to prepare n potions.

Examples
input

20 3 2
10 99
2 4 3
20 10 40
4 15
10 80

output

20

input

20 3 2
10 99
2 4 3
200 100 400
4 15
100 800

output

200

Note

In the first sample, the optimum answer is to use the second spell of the first type that costs 10 manapoints. Thus, the preparation time of each potion changes to 4 seconds. Also, Anton should use the second spell of the second type to instantly prepare 15 potions spending 80 manapoints. The total number of manapoints used is 10 + 80 = 90, and the preparation time is 4·5 = 20 seconds (15potions were prepared instantly, and the remaining 5 will take 4 seconds each).

In the second sample, Anton can't use any of the spells, so he just prepares 20 potions, spending 10 seconds on each of them and the answer is 20·10 = 200.

 枚举第一种 spell 用了哪一个(注意可能不用),然后观察到第二种 spell 的收益随着代价增大而增大,尽量选代价最大的,二分即可。

  1 #include<map>
2
3 #include<set>
4
5 #include<stack>
6
7 #include<cmath>
8
9 #include<queue>
10
11 #include<bitset>
12
13 #include<math.h>
14
15 #include<vector>
16
17 #include<string>
18
19 #include<stdio.h>
20
21 #include<cstring>
22
23 #include<iostream>
24
25 #include<algorithm>
26
27 #pragma comment(linker, "/STACK:102400000,102400000")
28
29 using namespace std;
30
31 typedef double db;
32
33 typedef long long ll;
34
35 typedef unsigned int uint;
36
37 typedef unsigned long long ull;
38
39 const db eps=1e-5;
40
41 const int N=2e5+10;
42
43 const int M=4e6+10;
44
45 const ll MOD=1000000007;
46
47 const int mod=1000000007;
48
49 const int MAX=1000000010;
50
51 const double pi=acos(-1.0);
52
53 ll a[N],b[N],c[N],d[N];
54
55 int main()
56
57 {
58
59 int i,m,k,l,r,mid;
60
61 ll n,x,s,t,ans;
62
63 scanf("%I64d%d%d", &n, &m, &k);
64
65 scanf("%I64d%I64d", &x, &s);
66
67 for (i=1;i<=m;i++) scanf("%I64d", &a[i]);
68
69 for (i=1;i<=m;i++) scanf("%I64d", &b[i]);
70
71 for (i=1;i<=k;i++) scanf("%I64d", &c[i]);
72
73 for (i=1;i<=k;i++) scanf("%I64d", &d[i]);
74
75
76 ans=n*x;
77
78 for (i=1;i<=k;i++)
79
80 if (d[i]<=s) ans=min(ans,x*max(0ll,n-c[i]));
81
82
83 for (i=1;i<=m;i++)
84
85 if (b[i]<=s&&a[i]<x) {
86
87 t=s-b[i];
88
89 if (t<d[1]) ans=min(ans,n*a[i]);
90
91 else {
92
93 l=1;r=k+1;mid=(l+r)>>1;
94
95 while (l+1<r)
96
97 if (d[mid]<=t) l=mid,mid=(l+r)>>1;
98
99 else r=mid,mid=(l+r)>>1;
100
101 ans=min(ans,max(0ll,n-c[l])*a[i]);
102
103 }
104
105 }
106
107 printf("%I64d\n", ans);
108
109 return 0;
110
111 }

CodeForce-734C Anton and Making Potions(贪心+二分)的更多相关文章

  1. C. Anton and Making Potions 贪心 + 二分

    http://codeforces.com/contest/734/problem/C 因为有两种操作,那么可以这样考虑, 1.都不执行,就是开始的答案是n * x 2.先执行第一个操作,然后就会得到 ...

  2. Codeforces 734C Anton and Making Potions(枚举+二分)

    题目链接:http://codeforces.com/problemset/problem/734/C 题目大意:要制作n个药,初始制作一个药的时间为x,魔力值为s,有两类咒语,第一类周瑜有m种,每种 ...

  3. Codeforces Round #379 (Div. 2) C. Anton and Making Potions 枚举+二分

    C. Anton and Making Potions 题目连接: http://codeforces.com/contest/734/problem/C Description Anton is p ...

  4. Codeforces 734C. Anton and Making Potions(二分)

    Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass ...

  5. Codeforces Round #379 (Div. 2) C. Anton and Making Potions 二分

    C. Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input ...

  6. Codeforces Round #379 (Div. 2) C. Anton and Making Potions —— 二分

    题目链接:http://codeforces.com/contest/734/problem/C C. Anton and Making Potions time limit per test 4 s ...

  7. 二分算法题目训练(三)——Anton and Making Potions详解

    codeforces734C——Anton and Making Potions详解 Anton and Making Potions 题目描述(google翻译) 安东正在玩一个非常有趣的电脑游戏, ...

  8. [二分] Codefoces Anton and Making Potions

    Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input sta ...

  9. Anton and Making Potions

    Anton and Making Potions time limit per test 4 seconds memory limit per test 256 megabytes input sta ...

随机推荐

  1. 【原创】在macOS Big Sur (Silicon M1, ARM)中配置ASP运行环境

    亲测有效,转载请附原文地址. 一,安装Parallels Desktop,注意选择支持ARM的版本. 二,注册 Windows Insider Preview Downloads 账号,通过以下链接下 ...

  2. Spring Boot 与 R2DBC 整合

    R2DBC 是 "Reactive Relational Database Connectivity"的简称.R2DBC 是一个 API 规范的倡议,声明对于访问关系型数据库驱动实 ...

  3. [C++]-unordered_map 映射

    unordered_map和map的区别请点击这里. 本文中的代码跟[C++]-map 映射中的代码仅仅是把定义的map类型数据定义成了unordered_map类型数据. 代码 #include&l ...

  4. tomcat及springboot实现Filter、Servlet、Listener

    tomcat实现: 核心类org.apache.catalina.startup.ContextConfig //支持注解 see:org.apache.catalina.deploy.WebXml ...

  5. IllegalArgumentException occurred while calling setter for property

    参考https://blog.csdn.net/qq_41192690/article/details/80659427 主码 是 integer类型的 就不要在写成这个样子了 把type=" ...

  6. MySQL-08-索引简介

    B树 基于不同的查找算法分类介绍 B*Tree B-tree B+Tree 在范围查询方面提供了更好的性能(> < >= <= like) 索引简介 索引作用 提供了类似于书中 ...

  7. systemd.service — 服务单元配置

    转载:http://www.jinbuguo.com/systemd/systemd.service.html 名称 systemd.service - 服务单元配置 大纲 service.servi ...

  8. [11 Go语言基础-可变参数函数]

    [11 Go语言基础-可变参数函数] 可变参数函数 什么是可变参数函数 可变参数函数是一种参数个数可变的函数. 语法 如果函数最后一个参数被记作 ...T ,这时函数可以接受任意个 T 类型参数作为最 ...

  9. Python - typing 模块 —— 常用类型提示

    前言 typing 是在 python 3.5 才有的模块 前置学习 Python 类型提示:https://www.cnblogs.com/poloyy/p/15145380.html 常用类型提示 ...

  10. MATLAB—M函数文件

    文章目录 一.M文件 1.脚本文件和函数文件的区别 二.M函数文件 1.创建方法 2.文件形式 2.函数形参及注释 3.函数的调用 4.函数变量 5.主函数.子函数 三.函数句柄 一.M文件 首先,要 ...