**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 60482 Accepted Submission(s): 20335

**

Problem 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

`while (scanf("%d %d",&m,&n)&&m!=-1)`

```scanf("%lf %lf",&a[i].a,&a[i].b);
if(a[i].b==0)
{
a[i].mods=DBL_MAX;
}
else
{
a[i].mods=a[i].a/a[i].b;
}```

```#include<iostream>
#include<stdio.h
#include<math.h>
#include<algorithm
#define DBL_MAX 1.7976931348623158e+308
using namespace std;
struct food
{
double a,b,mods;
};
food a[1001];
bool compare(food a,food b)
{
return a.mods<b.mods;
}
int main()
{
int m,n;
double ans;
while (scanf("%d %d",&m,&n)&&m!=-1)
{
for (int i = 0; i < n; i += 1)
{
scanf("%lf %lf",&a[i].a,&a[i].b);
if(a[i].b==0)
{
a[i].mods=DBL_MAX;
}
else
{
a[i].mods=a[i].a/a[i].b;
}
}
sort(a,a+n,compare);
ans=0.0;
while (n)
{
n--;
if (m>=a[n].b)
{
m-=a[n].b;
ans+=a[n].a;
}
else
{
ans+=m*a[n].mods;
m=0;
}
if(a[n].b!=0&&m==0)break;
}
printf("%.3lf\n",ans);
}
return 0;
}```

目录