【HDU 1009】FatMouse' Trade

题

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：给你m克猫粮，n个房间分别可以用猫粮以J[i]/F[i]的比例换取至多J[i]克咖啡豆，问最多能换取多少咖啡豆

分析：先排下序再贪心

#include <stdio.h>
#include <algorithm>
using namespace std;
int n,u;
double m,ans;
struct room
{
    double j,f,p;
} r[1005];
int cmp(room a,room b)
{
    return a.p>b.p;
}
int main()
{
    while(scanf("%lf%d",&m,&n)&&n!=-1&&m!=-1)
    {
        for(int i=0;i<n;i++)
        scanf("%lf%lf",&r[i].j,&r[i].f),
              r[i].p=r[i].j/r[i].f;
        sort(r,r+n,cmp);
        u=ans=0;
        while(m&&u<n)
        {
            if(r[u].f>m) ans+=m*r[u].p,m=0;
//剩下的猫粮不够把这个房间的咖啡豆换过来，那就能换多少换多少
            else ans+=r[u].j,m-=r[u].f;
//够的话，这间房子全部咖啡豆换过来
            u++;
        }
//        while(m>=r[u].f&&u<n) //换种写法
//        {
//            ans+=r[u].j;
//            m-=r[u].f;
//            u++;
//        }
//        if(u!=n) ans+=m*r[u].p;
        printf("%.3lf
",ans);

    }
    return 0;
}