Hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43381    Accepted Submission(s):
14499

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

 

Author

CHEN, Yue

 

贪心题：题目的意思是FatMouse有m磅的cat food，它想用这些cat food去换JavaBean，然后有n个房间，每个房间有相应的JavaBean和需要多少cat food。问m磅cat food最多可以换得多少JavaBean？

首先要将这些数据按照性价比（即JavaBean除以cat food 注意性价比有可能是小数）从大到小排序，然后问题就迎刃而解了。

 

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<iomanip>
 using namespace std;
 #define N 1005

 struct Trade{
     int J, F;       //J代表JavaBeans，F代表cat food
     double price;   //性价比
 }trade[N];

 int cmp(Trade a,Trade b)
 {
     return a.price>b.price;
 }
 int main()
 {
     double maximum;
     int m, n, i;
     while(cin>>m>>n)
     {
         maximum = ;
         if(m==- && n==-)
             break;
         maximum = ;
         for(i=; i<n; i++)
         {
             cin>>trade[i].J>>trade[i].F;
             trade[i].price = trade[i].J*1.0/trade[i].F;
         }
         sort(trade,trade+n,cmp);
         for(i=; i<n; i++)
         {
             if(m == )
                 break;
             else if(trade[i].F<=m)
             {
                 maximum += trade[i].J;
                 m -= trade[i].F;
             }
             else
             {
                 maximum += m*trade[i].price;
                 m = ;
             }
         }
         printf("%.3lf\n",maximum);
     }
     return ;
 }