HDU 1009 FatMouse' Trade

FatMouse' Trade

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

Total Submission(s): 80381    Accepted Submission(s): 27762


[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

 题目大意：一只大老鼠有M磅猫粮，然后他准备去跟猫做个交易以取得他喜欢的javabean。猫管理的仓库有n个房间，每个房间有J磅javabean而且总共需要f磅猫粮来换，请问怎样换就实惠。
赤裸裸的贪心....算出每个房间的单价（表达有问题.....），然后就买呗

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct Node{
double j,f,value;
}cat[1010];
bool cmp(Node a,Node b){
return a.value>b.value;
}
int main(){
int m,n;
while(~scanf("%d%d",&m,&n)){
if(m==-1&&n==-1) break;
for(int i=0;i<n;i++){
scanf("%lf%lf",&cat[i].j,&cat[i].f);
cat[i].value=cat[i].j/cat[i].f;
}
sort(cat,cat+n,cmp);
double sum=0;
for(int i=0;i<n;i++){
if(m>=cat[i].f){
m-=cat[i].f;
sum+=cat[i].j;
}
else{
sum+=cat[i].value*m;
break;
}
}
printf("%.3lf\n",sum);
}
}