hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62744    Accepted Submission(s): 21181

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

Source

ZJCPC2004

很简单的贪心，直接上代码了；

#include<iostream>#include<algorithm>#include<iomanip>using namespace std;struct room{double JavaBeans;double catfoods;double rate;};bool compare(room a,room b){return a.rate>b.rate;}int main(){int m,n;int i;room room[1001];while(cin>>m>>n){if(m==(-1)&&n==(-1)){return 0;}double s=0;for(i=0;i<n;i++){cin>>room[i].JavaBeans>>room[i].catfoods;room[i].rate=(room[i].JavaBeans/room[i].catfoods);}sort(room,room+n,compare);for(int k=0;k<n;k++){if(m>room[k].catfoods){s=s+room[k].JavaBeans;m=m-room[k].catfoods;}else{s=s+(double)m/room[k].catfoods*room[k].JavaBeans;break;}}cout.setf(ios::fixed);cout << setprecision(3) << s << endl;}return 0;}