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

简单贪心问题。先利用结构体对J和F进行排序。然后依次累加即可。

#include 
#include 
#include 
using namespace std;

struct Room
{
	int j, f;
};

Room room[1005];

bool cmp(const Room a, const Room b)
{
	return ((double)a.j / a.f > (double)b.j / b.f);
}

int main()
{
	int m, n;
	while (cin >> m >> n && m!=-1 && n!=-1)
	{
		int i;
		for (i = 0; i > room[i].j >> room[i].f;
		}
		sort(room , room + n, cmp);
		int jsum=0, fsum = 0;
		for (i = 0; i = m) { break; }
		}
		cout  m)
		{
			fsum -= room[i].f;
			jsum -= room[i].j;
			double javasum;
			javasum = jsum+(double)room[i].j *(double(m - fsum) / room[i].f);
			cout 
原文标题：HDOJ1009：FatMouse' Trade——简单贪心|落絮飞雁的个人网站
原文链接：https://www.luoxufeiyan.com/2015/03/29/hdoj1009/
授权协议：创作共用 署名-非商业性使用 2.5 中国大陆
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