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Hdu 1009 FatMouse' Trade

2024-07-18 07:55:06   224人阅读

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。
 
 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代表JavaBeans，F代表cat food10     double price;   //性价比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             else43             {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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