杭电-1009 FatMouse' Trade

最新推荐文章于 2021-02-26 01:08:20 发布

原创最新推荐文章于 2021-02-26 01:08:20 发布 · 313 阅读

0 ·

CC 4.0 BY-SA版权

OJ练习专栏收录该内容

218 篇文章

订阅专栏

此问题探讨了FatMouse如何通过最优交换策略，利用有限的猫粮换取仓库中最大数量的JavaBeans。涉及多个房间的不同交换比例，需找出最大收益。

FatMouse' Trade

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

Total Submission(s): 66325 Accepted Submission(s): 22518

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

Sample Output

  
   13.333
31.500

Author

CHEN, Yue

Source

ZJCPC2004

Recommend

JGShining

AC代码：

<pre name="code" class="cpp">#include <stdio.h>
#include <algorithm>
using namespace std;

struct Node
{
    double j,f,p;
} node[10000];

int cmp(Node x,Node y)
{
    return x.p>y.p;
}

int main()
{
    int m,n;
    while(~scanf("%d%d",&n,&m) && (m!=-1 || n!=-1))
    {
        double sum = 0,max = 0;
        int i,j;
        for(i = 0; i<m; i++)
        {
            scanf("%lf%lf",&node[i].j,&node[i].f);
            node[i].p = node[i].j/node[i].f;
        }
        sort(node,node+m,cmp);
        for(i = 0; i<m; i++)
        {
            if(n>=node[i].f)
            {
                sum+=node[i].j;
                n-=node[i].f;
            }
            else
            {
                sum+=node[i].p*n;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }

    return 0;
}