Hdu 1009 FatMouse' Trade 2016-05-05 23:02 86人阅读 评论(0) 收藏

FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 63198 Accepted Submission(s): 21342

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

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简单贪心题，按性价比排序后优先取大的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include<vector>
#include <map>
using namespace std;
#define inf 0x3f3f3f3f

struct node
{
    int a,b;
    double c;
} p[10000];

bool cmp(node a,node b)
{
return a.c>b.c;
}

int main()
{
    int m,n;
    while(~scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
            break;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&p[i].a,&p[i].b);
            p[i].c=p[i].a*1.0/p[i].b;
        }
        sort(p,p+n,cmp);
        double ans=0;
        for(int i=0;i<n;i++)
        {
            if(m>p[i].b)
            {
                ans+=p[i].a;
                m-=p[i].b;
            }
            else
            {
                ans+=m*p[i].c;
                break;
            }
        }
        printf("%.3f\n",ans);

    }
    return 0;
}