Hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43156 Accepted Submission(s): 14417

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

贪心.

贪心策略是根据价值比进行降序排序：J[i]/F[i];

代码如下：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6 #define MAX 1001
 7 struct node
 8 {
 9     int f;
10     int j;
11     double p;
12 };
13 struct node x[MAX];
14 int n,m;
15 bool cmp(struct node x,struct node y)
16 {
17     if(x.p==y.p) return x.f<y.f;
18     return x.p>y.p;
19 }
20 void out()
21 {
22     for(int i=0;i<n;i++)
23         cout<<x[i].p<<" ";
24     cout<<endl;
25 }
26 void init()
27 {
28     memset(x,0,sizeof(x));
29 }
30 void read()
31 {
32     int i;
33     for(i=0;i<n;i++)
34     {
35         scanf("%d %d",&x[i].j,&x[i].f);
36         x[i].p=(double)(x[i].j*1.0/(x[i].f*1.0));
37     }
38     sort(x,x+n,cmp);
39 }
40 void cal()
41 {
42     int i;
43     double res=m;
44     double ans=0;
45     for(i=0;i<n;i++)
46     {
47         if(res>=x[i].f)
48         {
49             res-=x[i].f;
50             ans+=x[i].j;
51         }
52         else
53         {
54             ans=ans+res*x[i].p;
55             res-=res;
56         }
57     }
58     printf("%.3f
",ans);
59 }
60 void solve()
61 {
62     init();
63     read();
64     cal();
65 }
66 
67 int main()
68 {
69     while(scanf("%d %d",&m,&n)!=EOF)
70     {
71         if(n==-1&&m==-1) break;
72         solve();
73     }
74     return 0;
75 }