B - 贪心 基础
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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
题目大意：老鼠贿赂猫，要偷吃东西，付出F[i]的猫粮可以吃到J[i]的javabean，同时每一个仓库里面的东西不一定全要，
你可以要任意百分比的东西，同时付出相应百分比的代价，问怎么选才能使老鼠吃到的javabean最多。
思路分析：首先区别于背包问题,背包问题中每一个物品是不能够拆分的，这也是为什么背包问题不能够贪心解决的原因，
用贪心做这道题，很显然，付出最少的代价得到最多的回报的策略是我们应该选择的，因为可以选择任意百分比，也就是说
最后的钱肯定可以花光，本道题的贪心策略就是选择性价比最高的物品优先购买，即value/cost值最大，以这个为标准从大
到小排序，每次把都选择性价比最高的仓库，就构成了正确答案。
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
using namespace std;
struct nod
{
double c;
double v;
double s;
};
const int maxn=1000+10;
nod java[maxn];
bool cmp(nod a,nod b)
{
return a.s>b.s;
}
int main()
{
int n,i;
double m,k;
while(scanf("%lf%d",&m,&n)&&(m!=-1||n!=-1))
{
double k=0;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&java[i].v,&java[i].c);
java[i].s=java[i].v/java[i].c;
}
sort(java,java+n,cmp);
for( i=0;i<=n-1;i++)
{
if(m>=java[i].c)
{
m-=java[i].c;
k+=java[i].v;
}
else
{
k+=(double)m/java[i].c*java[i].v;
break;
}
}
printf("%.3lf\n",k);
}
return 0;
}