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 <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct laoshu
{
int j;
int f;
double s;
};
bool cmp(laoshu stud1,laoshu stud2)
{
return stud1.s>stud2.s;
}
int main()
{
#ifdef CDZSC_OFFLINE
freopen("in.txt","r",stdin);
#endif
int i,m,n;
double sum;
laoshu stud[1010];
while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1)
{
sum=0;
for(i=0; i<n; i++)
{
scanf("%d%d",&stud[i].j,&stud[i].f);
stud[i].s=1.0*stud[i].j/stud[i].f;
}
sort(stud,stud+n,cmp);
for(i=0; i<n; i++)
{
if(m>=stud[i].f)
{
sum+=stud[i].j;
m-=stud[i].f;
}
else
{
sum+=stud[i].s*m;
break;
}
}
printf("%.3lf\n",sum);
}
return 0;
}