HDU(1009)FatMouse' Trade(贪心）

FatMouse' Trade

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

Total Submission(s): 66479 Accepted Submission(s): 22595



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

[align=left]Sample Output[/align]

13.333
31.500
题目描述：
一共有n个房子，每个房子里有老鼠喜欢吃的javabeans,但是每个房间里的javabeans的价格不一样。老鼠用m元，问m元最多可以卖多少javabeans，其中每个房间里的javabeans可以被分割。
[code]#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
struct luo
{
double x,y,z;
}lu[10000];
bool cmp(luo a,luo b)
{
return a.z>b.z;
}
int main()
{
int n,m;
while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1)
{double sum=0;
for(int i=0;i<n;i++)
{scanf("%lf%lf",&lu[i].x,&lu[i].y);
lu[i].z=lu[i].x/lu[i].y;｝//换算成单价
sort(lu,lu+n,cmp);
for(int i=0;i<n;i++)
{
if(m>=lu[i].y)
{sum+=lu[i].x;
m=m-lu[i].y;}
else
{
sum+=(m*lu[i].z);
break;
}
}
printf("%.3lf\n",sum);
}
}

[/code]