ZeptoLab Code Rush 2015 C. Om Nom and Candies [ 数学 ]
1 second
256 megabytes
standard input
standard output
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.
The single line contains five integers C, Hr, Hb, Wr, Wb (1 ≤ C, Hr, Hb, Wr, Wb ≤ 109).
Print a single integer — the maximum number of joy units that Om Nom can get.
10 3 5 2 3
16
In the sample test Om Nom can eat two candies of each type and thus get 16 joy units.
哎,做过一次,知道了最大公约数化简,最后枚举的时候忘了用大的枚举。。。
| 10596551 | 2015-04-05 05:09:39 | njczy2010 | C - Om Nom and Candies | GNU C++ | Accepted | 31 ms | 0 KB |
#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
#include <algorithm>
#include <queue>
#include <map> #define ll long long
int const N = ;
int const M = ;
int const INF = 0x3f3f3f3f;
ll const mod = ; using namespace std; ll c,hr,hb,wr,wb;
ll g;
ll ans;
ll ma;
ll lcm; ll gcd(ll a,ll b)
{
if(b==)
return a;
return gcd(b,a%b);
} void ini()
{
g=gcd(wr,wb);
if(wr<wb){
swap(wr,wb);
swap(hr,hb);
}
lcm=wr*wb/g;
ans=;
ma=;
} void solve()
{
if(c%lcm==){
ans=max(c/wr*hr,c/wb*hb);
return;
}
ll shang=c/lcm;
if(shang>=){
c-=lcm*(shang-);
ans=(shang-)*(max(lcm/wr*hr,lcm/wb*hb));
}
ll i;
for(i=;i*wr<=c;i++){
ma=max(ma,i*hr+(c-i*wr)/wb*hb);
}
ans+=ma;
} void out()
{
printf("%I64d\n",ans);
} int main()
{
//freopen("data.in","r",stdin);
//scanf("%d",&T);
//for(int cnt=1;cnt<=T;cnt++)
//while(T--)
while(scanf("%I64d%I64d%I64d%I64d%I64d",&c,&hr,&hb,&wr,&wb)!=EOF)
{
ini();
solve();
out();
}
}
ZeptoLab Code Rush 2015 C. Om Nom and Candies [ 数学 ]的更多相关文章
- 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 ...
- 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
- 小程序setData,视图层没有跟新
如果你完全符合微信介绍的setData使用说明的情况下,发现视图层没有更新,你可以看看我的这种情况. 使用setData的时候,修改的是data中一个对象的值,然后这个对象里面第一层不能含有 numb ...
- flex和box兼容性写法
display: -webkit-box; /* Chrome 4+, Safari 3.1, iOS Safari 3.2+ */ display: -moz-box; /* Firefox 17- ...
- Eric's并发用户数估算与Little定律的等价性
在国内性能测试的领域有一篇几乎被奉为大牛之作的经典文章,一个名叫Eric Man Wong 于2004年发表了名为<Method for Estimating the Number of Con ...
- vb6如何调用delphi DLL中的函数并返回字符串?
1,问题描述 最近发现vb6调用delphi DLL中的函数并返回字符串时出现问题,有时正常,有时出现?号,有时干脆导致VB程序退出 -- :: 将金额数字转化为可读的语音文字:1转化为1元 ???? ...
- 计算1至n的k次方的和
package com.ywx.count; import java.util.Scanner; /** * @author Vashon * date:20150410 * 题目:计算1至n的k次方 ...
- Android-Emulator使用
1.查看当前android支持的avd版本 2.创建Emulator avd: android create avd -n magicyu -t 2 -n后面接需要创建avd的名字,-t后面接需要创建 ...
- vue-cli下面的config/index.js注解 webpack.base.conf.js注解
config/indexjs详解上代码: 'use strict' // Template version: 1.3.1 // see http://vuejs-templates.github.io ...
- learnpythonthehardway EX41 相关
str.count() # str.count()方法用于统计字符串里某个字符出现的次数.可选参数为在字符串搜索的开始与结束位置. # str.count(sub, start= 0,end=len( ...
- Python3简明教程(四)—— 流程控制之分支
我们通过 if-else 语句来做决定,来改变程序运行的流程. if语句 语法如下: if expression: do this 如果表达式 expression 的值为真(不为零的任何值都为真), ...
- .net MVC下跨域Ajax请求(JSONP)
一.JSONP(JSON with Padding) 客户端: <script type="text/javascript"> function TestJsonp() ...