FatMouse' Trade ZOJ:2109

FatMouse' Trade

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

 

#include<bits/stdc++.h>
using namespace std;

struct node
{
	int f, j;
}nodes[1001];

bool compare(node x, node y) {
	return 1.0* x.j / x.f > 1.0 * y.j / y.f;
}

int main() {
	int m, n;
	while (cin >> m >> n && m != -1 && n != -1)
	{
		for (size_t i = 0; i < n; i++)
		{
			cin >> nodes[i].j >> nodes[i].f;
		}

		sort(nodes, nodes + n, compare);

		double sumJ(0);

		for (size_t i = 0; i < n; i++)
		{
			if (m - nodes[i].f >= 0)
			{
				m -= nodes[i].f;
				sumJ += nodes[i].j;
			}
			else
			{
				sumJ = sumJ + (1.0 * nodes[i].j / nodes[i].f) * m;
				break;
			}
		}
		cout << setiosflags(ios::fixed) << setprecision(3) << sumJ << endl;
	}
	//system("pause");
	return 0;
}