FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 93676 Accepted Submission(s): 32566

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

Author

CHEN, Yue

Source

ZJCPC2004

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题目意思：

一共有n个房子，每个房子里有老鼠喜欢吃的javabeans,但是每个房间里的javabeans的价格不一样。老鼠用m元，问m元最多可以卖多少javabeans，其中每个房间里的javabeans可以被分割。

分析：按照单价排序（价值除以重量），一直选择单价最大的的全部，当不能选择全部的时候，选择一部分

code：

#include<iostream>

#include<algorithm>

#include<stdio.h>

#include<queue>

#include<set>

#include<map>

#include<string>

#include<memory.h>

#include<math.h>

#define eps 1e-7

using namespace std;

#define max_v 1005

struct node

{

    double v,c;

}p[max_v];

bool cmp(node a,node b)

{

    return (a.v/a.c)>(b.v/b.c);//单价

}

int main()

{

    int n,m;

    double sum;

    int i;

    while(cin>>m>>n)

    {

        if(n==-&&m==-)

            break;

        for(i=;i<n;i++)

            cin>>p[i].v>>p[i].c;

        sort(p,p+n,cmp);

        sum=;

        for(i=;i<n;i++)

        {

            if(m>=p[i].c)

            {

                sum+=p[i].v;

                m-=p[i].c;

            }else

            {

                sum+=(m*(p[i].v/p[i].c));

                break;

            }

        }

        printf("%0.3lf\n",sum);

    }

    return ;

}

巴特西

HDU 1009 FatMouse' Trade（简单贪心）

FatMouse' Trade

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