ZOJ-2109FatMouse' Trade
/*
2109FatMouse' Trade
Time Limit: 1 Second Memory Limit: 32768 KB
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
*/
这是一道如此简单的题啊，最简单的贪心，可是却郁闷了我很久很久，是哪里郁闷我呢，是这么一组测试数据
0 3
7 0
0 4
5 0
只需考虑到这组测试数据里包含的陷阱就没问题了
#include <iostream>
#include <iomanip>
#include <algorithm>
using namespace std;
struct a{int j,f;double w;}p[1001];
bool cmp(const a m, const a n)
{
return m.w>n.w?1:0;
}
int main()
{
int m,n;
while(cin>>m>>n&&(m!=-1||n!=-1))
{
double total=0.0;
for(int i=0;i<n;++i)
{
cin>>p[i].j>>p[i].f;
if(p[i].f) p[i].w=p[i].j*1.0/p[i].f;
else p[i].w=9999; //food为0时，权值应无穷大，这里用比1000大的数就行
}
sort(p,p+n,cmp);
for(int i=0;i<n;++i)
if(m>=p[i].f)
{
total+=p[i].j;
m-=p[i].f;
}
else
{
total+=p[i].w*(m*1.0);
break;
}
cout<<fixed<<setprecision(3)<<total<<endl;
}
return 0;
}