//搞不清楚for循环真的害死人//贪心FatMouse's Trade------[NWPU][2018寒假作业][通用版]二E题

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

解题思路：

这道题思路跟前面做的两道贪心的题目想法类似。希望先交易j大f小的房间，所以计算j与f的比值，将其由大到小排序，从比值大的开始交易。

思路虽然简单，但是AC的道路却异常坎坷。

一开始WA的时候，我的第一反应是数据类型有问题，因为这道题有很多double型int型的数，又检查一遍之后发现还是WA。

然后我就觉得是sort的问题，事实证明我确实没有考虑求j和f比值时f为0的情况，但是改完之后还是WA。后来发现这个不用考虑，是我想多了。

于是找了网上AC的代码比对，发现最后应该用for循环而我用了while。我想了半天才明白，区别在于while循环只在m=0时退出，而for循环在m=0时和i>n时都会退出。就是说可能出现m还没用完但是所有房间都被买通了的情况。

我在while循环体最后面加了一个if(i>n) break;的语句后，还是WA，但是for循环的代码就AC了。想了半天这两个的区别，后来意识到for循环从一开始i=1时就执行条件表达式，而在我的while循环里面条件表达式是从i=2开始执行的。

也就是说，还存在n=0的测试数据！！！就是根本没有房间！！！

这谁想的到！！！我是陈独秀么？？？！！！

所以说，只有你想不到的，没有测不到的，还是不要给自己找不痛快，乖乖用for循环把。

#include<stdio.h>
#include<algorithm>
using namespace std;

double m;
int n;
struct node{
double j;
double f;
double s;
}room[1010];

bool cmp(node x,node y)
{
return x.s>y.s;
}

int main()
{
while(1)
{
scanf("%lf%d",&m,&n);
if(m==-1&&n==-1)
{
return 0;
}
for(int i=1;i<=n;i++)
{
scanf("%lf%lf",&room[i].j,&room[i].f);
room[i].s=room[i].j/room[i].f;
}
sort(room+1,room+n+1,cmp);
double ans=0;

4000
for(int i=1;i<=n;i++)
{
if(m>=room[i].f)
{
m-=room[i].f;
ans+=room[i].j;
}
else
{
ans+=m/room[i].f*room[i].j;
break;
}
}
printf("%.3lf\n",ans);
}
}