ZeptoLab Code Rush 2015 C. Om Nom and Candies 暴力
C. Om Nom and Candies
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/526/problem/C
Description
A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him to think a bit in order to enjoy candies the most. Would you succeed with the same task if you were on his place?
One day, when he came to his friend Evan, Om Nom didn't find him at home but he found two bags with candies. The first was full of blue candies and the second bag was full of red candies. Om Nom knows that each red candy weighs Wr grams and each blue candy weighs Wb grams. Eating a single red candy gives Om Nom Hr joy units and eating a single blue candy gives Om Nom Hb joy units.
Candies are the most important thing in the world, but on the other hand overeating is not good. Om Nom knows if he eats more than C grams of candies, he will get sick. Om Nom thinks that it isn't proper to leave candy leftovers, so he can only eat a whole candy. Om Nom is a great mathematician and he quickly determined how many candies of what type he should eat in order to get the maximum number of joy units. Can you repeat his achievement? You can assume that each bag contains more candies that Om Nom can eat.
Input
Output
Sample Input
10 3 5 2 3
Sample Output
HINT
题意
只有两个物品的多重背包问题
题解:
啊, 正解实在不会,只能乱搞
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/* int buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
printf("\n");
}
*/
//************************************************************************************** inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
} int main()
{
ll c,hr,hb,wr,wb;
ll ans=;
cin>>c>>hr>>hb>>wr>>wb;
if(wr==wb)
{
cout<<c/wr*max(hr,hb)<<endl;
return ;
}
if(wr<wb)
{
swap(wr,wb);
swap(hr,hb);
}
int time=;
for(int i=c/wr;i>=;i--)
{
time++;
ll sum=i*hr+(c-wr*i)/wb*hb;
ans=max(sum,ans);
if(time>)
break;
}
//cout<<ans<<endl;
swap(wr,wb);
swap(hr,hb);
time=;
for(int i=c/wr;i>=;i--)
{
time++;
ll sum=i*hr+(c-wr*i)/wb*hb;
ans=max(sum,ans);
if(time>)
break;
}
cout<<ans<<endl;
}
ZeptoLab Code Rush 2015 C. Om Nom and Candies 暴力的更多相关文章
- ZeptoLab Code Rush 2015 C. Om Nom and Candies [ 数学 ]
传送门 C. Om Nom and Candies time limit per test 1 second memory limit per test 256 megabytes input sta ...
- Codeforces - ZeptoLab Code Rush 2015 - D. Om Nom and Necklace:字符串
D. Om Nom and Necklace time limit per test 1 second memory limit per test 256 megabytes input standa ...
- ZeptoLab Code Rush 2015 B. Om Nom and Dark Park DFS
B. Om Nom and Dark Park Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/5 ...
- ZeptoLab Code Rush 2015 B. Om Nom and Dark Park
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who l ...
- Codeforces ZeptoLab Code Rush 2015 D.Om Nom and Necklace(kmp)
题目描述: 有一天,欧姆诺姆发现了一串长度为n的宝石串,上面有五颜六色的宝石.他决定摘取前面若干个宝石来做成一个漂亮的项链. 他对漂亮的项链是这样定义的,现在有一条项链S,当S=A+B+A+B+A+. ...
- CodeForces ZeptoLab Code Rush 2015
拖了好久的题解,想想还是补一下吧. A. King of Thieves 直接枚举起点和5个点之间的间距,进行判断即可. #include <bits/stdc++.h> using na ...
- Zepto Code Rush 2014 B - Om Nom and Spiders
注意题目给的是一个nxm的park,设元素为aij,元素aij 有4种可能U(上移),D(下移),L(左移),R(右移) 假设第i行第j列元素aij(注意元素的索引是从0开始的) 当aij为D时,此时 ...
- CF Zepto Code Rush 2014 B. Om Nom and Spiders
Om Nom and Spiders time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
- ZeptoLab Code Rush 2015 A. King of Thieves 暴力
A. King of Thieves Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/526/pr ...
随机推荐
- Nginx中worker_connections的问题
查看日志,有一个[warn]: 3660#0: 20000 worker_connections are more than open file resource limit: 1024 !! 原来安 ...
- C# 浅谈 接口(Interface)的作用
继承"基类"跟继承"接口"都能实现某些相同的功能,但有些接口能够完成的功能是只用基类无法实现的 1.接口用于描述一组类的公共方法/公共属性. 它不实现任何的方法 ...
- python基础-类的封装
封装:类中封装了公有属性和方法,对象封装了私有属性的值 class F1: def __init__(self,n): self.N=n print('F') class F2: def __init ...
- python基础-类的属性(类属性,实例属性,私有属性)
一:类的属性 类的属性分为:类属性(公有属性),实例属性和私有属性. 1)类属性(公有属性(静态字段): 类定义时直接指定的属性(不是在__init__方法中),可以通过类名直接访问属性,并且保存 ...
- Linux网络综合命令——IP
1.作用 ip是iproute2软件包里面的一个强大的网络配置工具,它能够替代一些传统的网络管理工具,例如ifconfig.route等,使用权限为超级用户.几乎所有的Linux发行版本都支持该命令. ...
- JS调用百度地图API标记地点
<!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content ...
- mysql数据库主从同步复制原理
MySQL的Replication(英文为复制)是一个多MySQL数据库做主从同步的方案,特点是异步复制,广泛用在各种对MySQL有更高性能.更高可靠性要求的场合.与之对应的是另一个同步技术是MySQ ...
- IPC对象的持续性
转载:http://book.51cto.com/art/201006/207275.htm <UNIX网络编程:第2版.第2卷,进程间通信>本书全面深入地讲解了各种进程间通信形式,包括消 ...
- 《精通Python设计模式》学习之抽象工厂
这种工厂模式用得少, 可能在游戏类的编程中用得比较多吧. 这个思路清晰一定要OK的. class Frog: def __init__(self, name): self.name = name de ...
- 转:40个Java集合面试问题和答案
转自牛客网:http://mp.weixin.qq.com/s?__biz=MjM5NDYxMzk1Nw==&mid=215319390&idx=1&sn=1ab621bc40 ...