杭电acm1009 FatMouse'Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 84996 Accepted Submission(s): 29492

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

Solution

贪心问题，主要就是先排序

Code

/** 
 * date:2017.11.14 
 * author:孟小德 
 * function:acm1099 
 *  FatMouse' Trade 
 */  
  
  
  
import java.util.*;  
import java.text.DecimalFormat;  
  
public class acm1009  
{  
    public static void main(String[] args)  
    {  
        Scanner input = new Scanner(System.in);  
        DecimalFormat df = new DecimalFormat("0.000");  
  
        int catf_num;  
        int room_num;  
  
          
        ArrayList result = new ArrayList();  
        while ((catf_num = input.nextInt()) != -1 && (room_num = input.nextInt()) != -1)  
        {  
            double[][] source = new double[room_num][2];  
            double[] persent = new double[room_num];  
            for (int i=0;i=0;i--)  
            {  
                // int n = map.get(persent[i]);  
                if (source[i][1] <= catf_num)  
                {  
                    getNum += source[i][0];  
                    catf_num -= source[i][1];  
                }  
                else  
                {  
                    getNum += catf_num * persent[i];  
                    break;  
                }  
            }  
  
            System.out.printf("%.3f",getNum);  
            System.out.println();  
  
        }  
  
  
    }  
  
    //对输入的resource进行排序  
    public static void sortArray(int[][] source,double[] persent)  
    {  
        boolean flag = true;  
        Double temp;  
        int[] tempArray = new int[2];  
        for (int i = 0;i persent[j+1])  
                {  
                    temp = persent[j];  
                    persent[j] = persent[j+1];  
                    persent[j+1] = temp;  
  
                    tempArray[0] = source[j][0];  
                    tempArray[1] = source[j][1];  
  
                    source[j][0] = source[j+1][0];  
                    source[j][1] = source[j+1][1];  
  
                    source[j+1][0] = tempArray[0];  
                    source[j+1][1] = tempArray[1];  
  
                    flag = true;  
                }  
            }  
        }  
    }  
  
  
    static void sort(double[] a,double[][] c) {  
        int len = a.length;  
        int low = 0,high = len - 1;  
        quickSort(a,c, low, high);  
    }  
  
    static void quickSort(double[] a, double[][] c,int l ,int h){  
        if(l>=h){  
            return;  
        }  
        int low = l;  
        int high = h;  
        double k = a[low];  
        double k2 = c[low][0];  
        double k3 = c[low][1];  
        while(low< high){  
            //  
            while(high>low&&a[high]>=k){//寻找元素右边比其小的  
                high --;  
            }  
            a[low] = a[high];//进行交换，K指向high  
            c[low][0] = c[high][0];  
            c[low][1] = c[high][1];  
            while(low