HDU 1009 FatMouse' Trade

FatMouse' Trade

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

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

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN = 1010;
struct node
{
double j,f;
double r;
} a[MAXN];
bool cmp(node a,node b)
{
return a.r  >  b.r;
}
int main()
{
int N;
double M;
double ans;
while(scanf("%lf%d",&M,&N))
{
if(M==-1&&N==-1) break;
for(int i=0; i<N; i++)
{
scanf("%lf%lf",&a[i].j,&a[i].f);
a[i].r=(double)a[i].j/a[i].f;
}
sort(a,a+N,cmp);
ans=0;
for(int i=0; i<N; i++)
{
if(M>=a[i].f)
{
ans+=a[i].j;
M-=a[i].f;
}
else
{
ans+=(a[i].j/a[i].f)*M;
break;
}
}
printf("%.3lf\n",ans);
}
return 0;
}