題目：

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5092 Accepted Submission(s): 1530 Problem DescriptionFatMouse 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.
InputThe 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.
OutputFor 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

AuthorCHEN, Yue SourceZJCPC2004 RecommendJGShining

題目大意：

老鼠准備了M磅貓食，准備拿這些貓食跟貓交換自己喜歡的食物。有N個房間，每個房間裡面都有食物。你可以得到J[i]但你需要付出F[i]的貓食。要你計算你有M磅貓食可以獲得最多食物的重量。而且這裡可以不必每一組都全換，可以按比例換。。。例如最後你只剩5塊錢貓食。但是目前的一個選擇是話10塊貓食就能換取6個糧食。那麼這時候你用5塊貓食就能換取3個糧食

題目分析：

這是一道簡單的貪心題。對於這種獲得收入的同時需要付出一些的情況下，計算最大收入。這種題一般是根據

收入和付出的比例來排一下序。然後根據這個比例從高到低進行選擇

代碼如下：

/*
 * a.cpp
 *
 *  Created on: 2015年1月28日
 *      Author: Administrator
 */

#include 
#include 
#include 

using namespace std;

const int maxn = 10005;

struct W {
	double get;//收入
	double pay;//付出的貓食
	double ave;//收入/付出比
} w[maxn];

bool cmp(W a, W b) {
	if (a.ave > b.ave) {
		return true;
	}

	return false;
}

int main() {
	int n;
	double m;
	while (scanf("%lf%d", &m, &n), n != -1 && m != -1) {
		int i;
		for (i = 0; i < n; ++i) {
			scanf("%lf %lf", &w[i].get, &w[i].pay);
			w[i].ave = w[i].get / w[i].pay;
		}

		sort(w, w + n, cmp);//貪心,對w按照get/pay進行降序排序

		double sum = 0;
		i = 0;
//		while(m >= 0){//不知道為什麼這種寫法就是不行.
//			if (m >= w[i].pay) {
//							sum = sum + w[i].get;
//							m = m - w[i].pay;
//						} else {
//							sum = sum + w[i].ave * m;
//							break;
//						}
//			i++;
//		}

		for (i = 0; i <= n - 1; i++) {
			if (m >= w[i].pay) {//如果當前剩余的貓食還足夠的話
				sum = sum + w[i].get;//那就把那個房間的糧食全部買下
				m = m - w[i].pay;//並且手上見去相應的貓食
			} else {//如果現在手上的貓食已經不夠
				sum = sum + w[i].ave * m;//那麼就按比例拿去一定的貓食
				break;
			}
		}

		printf("%.3lf\n", sum);
	}

	return 0;
}