链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1109
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
翻译：
从前有仅仅肥肥的老鼠。他叫FatMouse，他就像人类的恐怖分子跟敌人交易军火一样，猥琐的他准备了M磅猫食，准备与守卫仓库的大猫们进行交易，仓库里有他最爱吃的食物Javabean。
仓库里有N个房间，第i间房间里有J[i]磅Javabean且须要F[i]磅猫食进行交换，FatMouse不必吧每一个房间里的Javabean所实用于交易，相反。他可以付给大猫F[i]*a%磅猫食，从而换的J[i]*a%磅的Javabean。当中，a是一个实数，如今他给你布置一个家庭作业，请你告诉他他最多可以获得多少磅Javabean。
输入描写叙述：
输入包括多组測试数据，每组測试数据的开头一行是两个非负整数M, N.接下来的N行中，每行包括两个非负整数J[i]和F[i]，最后一组測试数据是两个-1。全部的整数的值不糊超过1000。
输出描写叙述：
对于每组測试数据，在一行上打印出一个3位小数的实数，这个实数是FatMouse可以交易到的最大数量的Javabean.
解题思路：
本题要求输出最大交易量。并保留三位小数。这样，我们使用J[i]除以F[i]就得到了a，那么，交易的时候，为了获得最多的Javabean，要先交易a大的。这样就确保了能交易到最多的Javabean.
把数据读入结构体中，再将结构体作为向量的元素，再按a由大到小的顺序给向量排序，然后依次进行计算。这样的题目属于背包类的题目！
(dp + 贪心)
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
#define MAXN 10005
#define RST(N)memset(N, 0, sizeof(N))
#include <algorithm>
using namespace std;
typedef struct Mouse_ {
double J, F;
double a;
}Mouse;
int n, m;
vector <Mouse> v;
vector <Mouse> ::iterator it;
bool cmp(const Mouse m1, const Mouse m2)
{
if(m1.a != m2.a) return m1.a > m2.a;
else return m1.F < m2.F;
}
int main()
{
while(~scanf("%d %d", &n, &m)) {
if(n == -1 && m == -1) break;
Mouse mouse;
v.clear();
for(int i=0; i<m; i++) {
scanf("%lf %lf", &mouse.J, &mouse.F);
mouse.a = mouse.J/mouse.F;
v.push_back(mouse);
}
sort(v.begin(), v.end(), cmp);
double sum = 0;
for(int i=0; i<v.size(); i++) {
if(n > v[i].F) {
sum += v[i].J;
n -= v[i].F;
}else {
sum += n*v[i].a;
break;
}
}
printf("%.3lf\n", sum);
}
return 0;
}
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ZOJ 2109 FatMouse' Trade （背包 dp + 贪婪）