Hdu - 1009 - FatMouse' Trade

上题目

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33316    Accepted Submission(s): 10813

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

 

 

      题意就是有m单位猫食，n堆的豆，每堆豆有一个余量和要这一堆豆的代价，每一堆豆可以只要一部分，那么付出的代价就按这一部分占这一堆豆的余量的百分比算。问用这m单位猫食可以最多得到多少豆。

　   贪心的基础题，白书上也有类似的题目。解法就是得出每一堆豆的单价(余量/代价)，先取完单价最高的豆，然后再取第二的···直到猫食用完。

 

上代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define MAX 1000+10
using namespace std;

typedef struct
{
    double J,F;
    double S;
}price;

bool cmp(price x,price y) {return x.S>y.S;}

price p[MAX];

int main()
{
    int i,m,n;
    double sum;
    //freopen("data.txt","r",stdin);
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        if(m==-1 && n==-1) break;
        memset(p,0,sizeof(p));
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&p[i].J,&p[i].F);
            p[i].S=p[i].J/p[i].F;
        }
        sort(p,p+n,cmp);
        sum=0;
        for(i=0;i<n;i++)
        {
            if(m>=p[i].F) {m-=p[i].F;sum+=p[i].J;}
            else {sum+=p[i].J*m/p[i].F;break;}
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}