题干：
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
题目大意：
肥鼠准备了 M 磅的猫粮，准备和看管仓库的猫交易，仓库里装有他最喜爱的食物 Java 豆。
仓库有 N 个房间。第 i 间房包含了 J[i] 磅的 Java 豆，需要 F[i] 磅的猫粮。肥鼠不必为了房间中的所有 Java 豆而交易，相反，他可以支付 F[i] * a% 磅的猫粮去交换得到 J[i] * a% 磅的 Java 豆。这里，a 表示一个实数。
现在他将这项任务分配给了你：请告诉他，能够获得的 Java 豆的最大值是多少。
注意这个‘%’是百分之！！不是取模！！！傻子了吧、、、
解题报告：
赤果果的贪心，，排序后先买最值得买的，直到钱花光了。
AC代码：
using namespace std;
int n;
double m,ans;
struct Node {
int j,f;
} node[1005];
bool cmp(Node a,Node b) {
return 1.0*a.j/a.f > 1.0*b.j/b.f;
}
int main()
{
while(cin>>m>>n) {
if(n == -1 && m == -1) break;
for(int i = 1; i<=n; i++) {
scanf("%d%d",&node[i].j,&node[i].f);
}
ans=0;
sort(node+1,node+n+1,cmp);
for(int i = 1; i<=n; i++) {
if(m >= node[i].f) {
m-=node[i].f;
ans+=node[i].j;
}
else {
ans += (m/node[i].f)*node[i].j;
break;
}
}
printf("%.3lf\n",ans);
}
return 0 ;
}
