FatMouse' Trade（杭电1009）

FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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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

//本题是简单贪心问题。意思是老鼠有m榜猫粮，通过给猫粮东西data[i].b能够兑换data[i].a，求最多能够兑换多少食物。

//仅仅需将data[i].a/data[i].b的值从大到小排序就可以求解。

#include<stdio.h>
struct st
{
    double a;
    double b;
    double c;
}data[1000];
int main()
{
    int i,j,m,n;
    struct st data[1000],t;
    while(scanf("%d %d",&m,&n)&&(m!=-1||n!=-1))
    {
        double sum=0.000;
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&data[i].a,&data[i].b);
        }
        for(i=0;i<n;i++)
        {
            for( j=i+1;j<n;j++)
            {
                if((data[i].a/data[i].b)<data[j].a/data[j].b)
                {
                t=data[i];
                data[i]=data[j];
                data[j]=t;}
            }
        }
        for(i=0;i<n;i++)
        {
            if(m-data[i].b>=0.001)
            {
                sum+=data[i].a;
                m-=data[i].b;
            }
            else
            {
                sum=sum+m*data[i].a/data[i].b;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}