HDU-ACM1009--贪心算法

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63073    Accepted Submission(s): 21296

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
   
   
   
   

import java.util.Scanner;

class Main{
    public static void main(String[] args) {
        Scanner input=new Scanner(System.in);
        while (input.hasNext()){
        	double m=input.nextDouble();
        	int n=input.nextInt();
        	if (m==-1&&n==-1){
        		return;
        	}
        	double j[]=new double[n];
        	double f[]=new double[n];
        	for (int i=0;i<n;i++){
        		j[i]=input.nextDouble();
        		f[i]=input.nextDouble();
        	}
        	for (int i=0;i<n-1;i++){
        		for (int k=0;k<n-1-i;k++){
        			if (j[k]/f[k]<j[k+1]/f[k+1]){
            			double t=0;
            			t=j[k];j[k]=j[k+1];j[k+1]=t;
            			t=f[k];f[k]=f[k+1];f[k+1]=t;
            		}
        		}
        	}
        	double sum=0;
        	if (m!=0){
        		for (int i=0;i<n;i++){
            		if (m<f[i]){
            			sum+=m*j[i]/f[i];
            			break;
            		}else {
            			sum+=j[i];
            			m-=f[i];
            			if (m==0){
            				break;
            			}
            		}
            	}
        		System.out.printf("%.3f",sum);
        		System.out.println();
        	}else {
        		System.out.println("0.000");
        	}
        }
    }
}