FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 63198 Accepted Submission(s): 21342

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.

The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1’s. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

5 3

7 2

4 3

5 2

20 3

25 18

24 15

15 10

-1 -1

Sample Output

13.333

31.500


简单贪心题,按性价比排序后优先取大的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include<vector>
#include <map>
using namespace std;
#define inf 0x3f3f3f3f struct node
{
int a,b;
double c;
} p[10000]; bool cmp(node a,node b)
{
return a.c>b.c;
} int main()
{
int m,n;
while(~scanf("%d%d",&m,&n))
{
if(m==-1&&n==-1)
break;
for(int i=0; i<n; i++)
{
scanf("%d%d",&p[i].a,&p[i].b);
p[i].c=p[i].a*1.0/p[i].b;
}
sort(p,p+n,cmp);
double ans=0;
for(int i=0;i<n;i++)
{
if(m>p[i].b)
{
ans+=p[i].a;
m-=p[i].b;
}
else
{
ans+=m*p[i].c;
break;
}
}
printf("%.3f\n",ans); }
return 0;
}

Hdu 1009 FatMouse' Trade 2016-05-05 23:02 86人阅读 评论(0) 收藏的更多相关文章

  1. PAT甲 1009. Product of Polynomials (25) 2016-09-09 23:02 96人阅读 评论(0) 收藏

    1009. Product of Polynomials (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...

  2. ZOJ2478 Encoding 2017-04-18 23:02 43人阅读 评论(0) 收藏

    Encoding Time Limit: 2 Seconds      Memory Limit: 65536 KB Given a string containing only 'A' - 'Z', ...

  3. POJ3281 Dining 2017-02-11 23:02 44人阅读 评论(0) 收藏

    Dining Description Cows are such finicky eaters. Each cow has a preference for certain foods and dri ...

  4. ImageView一例 分类: H1_ANDROID 2013-10-30 23:02 1812人阅读 评论(0) 收藏

    参考自<疯狂android讲义>2.4节 效果如下: 当点击图上某点时,将之附近放大至下图. 布局文件: <LinearLayout xmlns:android="http ...

  5. HDU 2101 A + B Problem Too 分类: ACM 2015-06-16 23:57 18人阅读 评论(0) 收藏

    A + B Problem Too Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  6. Uniform Generator 分类: HDU 2015-06-19 23:26 11人阅读 评论(0) 收藏

    Uniform Generator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...

  7. HDU 2040 亲和数 [补] 分类: ACM 2015-06-25 23:10 10人阅读 评论(0) 收藏

    今天和昨天都没有做题,昨天是因为复习太累后面忘了,今天也是上午考毛概,下午又忙着复习计算机图形学,晚上也是忘了结果打了暗黑3,把暗黑3 打通关了,以后都不会玩太多游戏了,争取明天做3题把题目补上,拖越 ...

  8. HDU 2034 人见人爱A-B 分类: ACM 2015-06-23 23:42 9人阅读 评论(0) 收藏

    人见人爱A-B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

  9. HDU 2035 人见人爱A^B 分类: ACM 2015-06-22 23:54 9人阅读 评论(0) 收藏

    人见人爱A^B Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Su ...

随机推荐

  1. 778A String Game

    A. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard inpu ...

  2. c#while循环注意continue的地方

    在使用while 时发现一个很大的问题,continue最好慎用! private void do() { int i = 0; while (true) { //continue;//绝对的死循环 ...

  3. Raw Socket(原始套接字)实现Sniffer(嗅探)

    参考资料: https://www.xuebuyuan.com/3190946.html https://blog.csdn.net/zxygww/article/details/52093308 i ...

  4. The server time zone value 'Öйú±ê׼ʱ¼ä' is unrecognized or represents more than one time zone问题解决

    从错误即可知道是时区的错误,因此只要将时区设置为你当前系统时区即可 因此使用root用户登录mysql,按照如下图所示操作即可. 把时区设置为所在地时区(即东八区的时区)后,再连接数据库就可以了

  5. TI and RI

    https://blog.csdn.net/qq_27977257/article/details/70677661 51单片机的串口,是个全双工的串口,发送数据的同时,还可以接收数据.当串行发送完毕 ...

  6. Python之路(第十篇)迭代器协议、for循环机制、三元运算、列表解析式、生成器

    一.迭代器协议 a迭代的含义 迭代器即迭代的工具,那什么是迭代呢? #迭代是一个重复的过程,每次重复即一次迭代,并且每次迭代的结果都是下一次迭代的初始值 b为何要有迭代器? 对于序列类型:字符串.列表 ...

  7. Eigen中的map

    Map类用于通过C++中普通的连续指针或者数组 (raw C/C++ arrays)来构造Eigen里的Matrix类,这就好比Eigen里的Matrix类的数据和raw C++array 共享了一片 ...

  8. Asterisk的type类型和身份认证

    Asterisk的type类型和身份认证 转载:http://zeevli.blog.163.com/blog/static/119591610201111745012380/ 在Asterisk中对 ...

  9. Angular学习笔记:Angular CLI

    定义 Angular CLI:The Angular CLI is a command line interface tool that can create a project, add files ...

  10. SparkStreaming--reduceByKeyAndWindow

    1.reduceByKeyAndWindow(_+_,Seconds(3), Seconds(2))     可以看到我们定义的window窗口大小Seconds(3s) ,是指每2s滑动时,需要统计 ...