HDOJ1009-FatMouse' Trade(贪心)

Posted 2020-09-21

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

 

Code:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define CLR(x , v)      memset(x , v , sizeof(x))
 5 static const int MAXN = 1e5 + 10;
 6 
 7 struct Node
 8 {
 9     int v , c;
10     double av;
11     Node(int _v = 0 , int _c = 0 , double _av = 0.0):v(_v) , c(_c) , av(_av) {}
12 };
13 Node data[MAXN];
14 bool cmp(Node a , Node b)
15 {
16     return a.av > b.av;
17 }
18 int n , m;
19 int main()
20 {
21     while(~scanf("%d%d" , &m , &n))
22     {
23         if(m == -1 && n == -1)
24             return 0;
25         CLR(data , 0);
26         for(int i = 1 ; i <= n ; ++i)
27         {
28             int a , b;
29             scanf("%d%d" , &a , &b);
30             data[i] = {a , b , a * 1.0 / b};
31         }
32 
33         sort(data + 1 , data + 1 + n , cmp);
34 
35         double ans = 0;
36 
37         for(int i = 1 ; i <= n ; ++i)
38         {
39             if(m > data[i].c)
40                 ans += data[i].v , m -= data[i].c;
41             else
42             {
43                 ans += (data[i].av * m);
44                 break;
45             }
46         }
47 
48         printf("%.3f\\n" , ans);
49     }
50 }

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