FatMouse' Trade

A - FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 37 24 35 220 325 1824 1515 10-1 -1 

 

Sample Output

 13.33331.500 
参考代码：

 #include<stdio.h>
#include<algorithm>
using namespace std;
typedef struct {
double f;
double j;
double av;
}note;
note a[1010];
bool cmp(note x,note y)
{
return x.av>y.av;
}
int main()
{
int n,i,j;
double m,s;
while(scanf("%lf%d",&m,&n),m!=-1||n!=-1)
{
s=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&a[i].f,&a[i].j);
a[i].av=a[i].f/a[i].j;
}
sort(a,a+n,cmp);
for(i=0;i<n;i++)
{
if(m>a[i].j)
{
s+=a[i].f;
m=m-a[i].j;
}
else
{
for(j=0;j<m;j++)
{
s+=a[i].av;
}
break;
}
}
printf("%.3lf\n",s);
}
return 0;
}