HDU 1009 FatMouse' Trade

题目链接：HDU 1009 FatMouse' Trade

题目大意：
给你\(m\)克猫粮，\(n\)个房间分别可以用猫粮以\(J[i]/F[i]\)的比例换取至多\(J[i]\)克咖啡豆，问最多能换取多少咖啡豆。

题解：
按照性价比排序，总取性价比最大的。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int n, m;
struct Room {
	double j, f, rate;
	bool operator < (const Room &x) const {
		return rate > x.rate;
	}
} room[1010];

int main() {
	while (~scanf("%d%d", &m, &n)) {
		if (n == -1 && m == -1) {
			break;
		}
		for (int i = 1; i <= n; ++i) {
			scanf("%lf%lf", &room[i].j, &room[i].f);
			room[i].rate = room[i].j / room[i].f;
		}
		sort(room + 1, room + n + 1);
		double ans = 0;
		for (int i = 1; i <= n; ++i) {
			if (room[i].f <= m) {
				ans += room[i].j;
				m -= room[i].f;
			} else {
				ans += room[i].rate * m;
				break;
			}
		}
		printf("%.3lf\n", ans);
	}
	return 0;
}