HDU 1009 FatMouse' Trade

Posted 2020-06-09

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 59851    Accepted Submission(s): 20095

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

 

 

Source

ZJCPC2004

 

 

Recommend

JGShining

 

 

使用格式输入输出的格式说明符%d,%f,%lf等，应当注意与int，float，double等类型的对应关系。

 

 

 

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int MAX=1050;
 6 struct Room{
 7     double j;
 8     double f;
 9     double w;
10 }r[MAX];
11 
12 bool compare(const Room &r1,const Room &r2)
13 {
14     if(r1.w>r2.w)
15         return true;
16     else
17         return false;
18 }
19 
20 int main()
21 {
22     double m,ans,jv,fv;
23     int n;
24     while(scanf("%lf%d",&m,&n)==2&&n!=-1)
25     {
26         ans=0;
27         for(int i=0;i<n;i++)
28         {
29             scanf("%lf%lf",&jv,&fv);
30             r[i].j=jv;
31             r[i].f=fv;
32             r[i].w=jv/fv;
33         }
34         sort(r,r+n,compare);
35 
36         for(int i=0;i<n;i++)
37         {
38             if(m>=r[i].f)
39             {
40                 m-=r[i].f;
41                 ans+=r[i].j;
42             
43             }
44             else
45             {
46                 ans+=r[i].w*m;
47                 break;
48             }
49         }
50         printf("%.3lf\n",ans);
51     }
52 }