hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 77049    Accepted Submission(s): 26462

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
解题思想：贪心算法 先求单价，然后排个序就行了。
java代码
import java.util.Scanner;public class hud1009 {/** * @param args */public static void main(String[] args) {// TODO Auto-generated method stubScanner scan=new Scanner(System.in);int m,n;double sum;while(((m=scan.nextInt())!=-1)&&((n=scan.nextInt())!=-1)){sum=0;double j[]=new double[n];double f[]=new double[n];double p[]=new double[n];for(int i=0;i<n;i++){double a=scan.nextDouble();double b=scan.nextDouble();j[i]=a;f[i]=b;p[i]=a/b;}sort(j,f,p,n);for(int i=0;i<n;i++){if(m>=f[i]){sum+=j[i];m-=f[i];}else{sum+=p[i]*m;break;}}System.out.printf("%.3f",sum);System.out.println();}}private static void sort(double[] j, double[] f, double[] p, int n) {// TODO Auto-generated method stubdouble temp;double temp1;double temp2;for(int i=0;i<n-1;i++){for(int k=0;k<n-1-i;k++){if(p[k]<p[k+1]){temp=p[k];p[k]=p[k+1];p[k+1]=temp;temp1=f[k];f[k]=f[k+1];f[k+1]=temp1;temp2=j[k];j[k]=j[k+1];j[k+1]=temp2;}}}}}