HDU1009_FatMouse' Trade【贪心】【水题】

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

题目大意：有N个房间，每一个房间存有FatMouse喜欢吃的食物。可是每一个房间

的食物都须要用对应的猫粮去换。

FatMouse 有M磅的猫粮，为它最多能换到多

少的食物。

思路：贪心方法。用结构体存每间房间的食物量和所需猫粮量。

按食物的单位价格（

即食物/猫粮的大小）进行排列，每次选单位价格最小的购买，知道M磅猫粮用完

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct warehouse
{
    double j;
    double f;
}a[1100];
bool cmp(warehouse a,warehouse b)
{
    return a.f/a.g < b.f/b.g;
}
int main()
{
    int N;
    double M;
    while(~scanf("%lf%d",&M,&N)&& (M!=-1||N!=-1))
    {
        memset(a,0,sizeof(a));
        for(int i = 0; i < N; i++)
        {
            scanf("%lf%lf",&a[i].j,&a[i].f);
        }
        sort(a,a+N,cmp);
        double sum = 0;
        for(int i = 0; i < N; i++)
        {
            if(M <= 0.000001)
                break;
            if(M >= a[i].f)
            {
                sum += a[i].j;
                M -= a[i].f;
            }
            else
            {
                sum += M*a[i].j/a[i].f;
                M = 0;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}