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

 

貪心題：題目的意思是FatMouse有m磅的cat food，它想用這些cat food去換JavaBean，而後有n個房間，每一個房間有相應的JavaBean和須要多少cat food。問m磅cat food最多能夠換得多少JavaBean？

首先要將這些數據按照性價比（即JavaBean除以cat food 注意性價比有多是小數）從大到小排序，而後問題就迎刃而解了。

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<iomanip>
 5 using namespace std;
 6 #define N 1005
 7 
 8 struct Trade{
 9     int J, F;       //J表明JavaBeans，F表明cat food
10     double price;   //性價比
11 }trade[N];
12 
13 int cmp(Trade a,Trade b)
14 {
15     return a.price>b.price;
16 }
17 int main()
18 {
19     double maximum;
20     int m, n, i;
21     while(cin>>m>>n)
22     {
23         maximum = 0;
24         if(m==-1 && n==-1)
25             break;
26         maximum = 0;
27         for(i=0; i<n; i++)
28         {
29             cin>>trade[i].J>>trade[i].F;
30             trade[i].price = trade[i].J*1.0/trade[i].F;
31         }
32         sort(trade,trade+n,cmp);
33         for(i=0; i<n; i++)
34         {
35             if(m == 0)
36                 break;
37             else if(trade[i].F<=m)
38             {
39                 maximum += trade[i].J;
40                 m -= trade[i].F;
41             }
42             else
43             {
44                 maximum += m*trade[i].price;
45                 m = 0;
46             }
47         }
48         printf("%.3lf\n",maximum);
49     }
50     return 0;
51 }

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