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 }
HDOJ1009-FatMouse' Trade(贪心)