FatMouse' Trade
http://acm.zju.edu.cn/show_problem.php?pid=2109
Time limit: 1 Seconds
Memory limit: 32768K
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.
1. 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.
2. 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.
3. Sample Input
{{{5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1}}}
4. Sample Output
{{{13.333 31.500}}}
1 /*Written by czk*/
2 #include <iostream>
3 #include <iomanip>
4 #include <queue>
5 using namespace std;
6 struct room{
7 int java_bean;
8 int cat_food;
9 bool operator<(const room &r) const{
10 return static_cast<double>(java_bean) / cat_food < static_cast<double>(r.java_bean) / r.cat_food;
11 }
12 };
13
14 int main() {
15 while(1) {
16 int m, n;
17 cin >> m >> n;
18 if (m == -1 && n == -1) break;
19 priority_queue<room> rooms;
20 for(int i = 0; i < n; i++) {
21 room r;
22 cin >> r.java_bean >> r.cat_food;
23 rooms.push(r);
24 }
25 double total = 0.0;
26 while(!rooms.empty() && m > 0) {
27 if(m>rooms.top().cat_food) {
28 total += rooms.top().java_bean;
29 m -= rooms.top().cat_food;
30 } else {
31 total += static_cast<double>(rooms.top().java_bean) / rooms.top().cat_food * m;
32 m = 0;
33 }
34 rooms.pop();
35 }
36 cout << fixed<<setprecision(3) << total << endl;
37 }
38
39 }