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.
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.
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.
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
#include<iostream>
#include<algorithm>
using namespace std;
struct room
{
int catfood;
int mousefood;
double danwei;
};
int cmp(room a,room b)
{
return a.danwei>b.danwei;
}
int main()
{
int M,N,i;
room a[1000];
while(scanf("%d%d",&M,&N))
{
double sum=0;
if(M==-1 && N==-1)
break;
for(i=0;i<N;i++)
{
cin>>a[i].mousefood>>a[i].catfood;
}
for(i=0;i<N;i++)
a[i].danwei=a[i].mousefood*1.0/a[i].catfood;//单位猫食,获得的物品量
sort(a,a+N,cmp);//降序排列
i=0;
while(i<N && a[i].catfood<=M)
{
sum=sum+a[i].mousefood;
M=M-a[i].catfood;
i++;
}
if(i==N)
{
printf("%.3lf\n",sum);//准备的猫食可能有剩余
}
else
{
sum=sum+a[i].mousefood*M*1.0/a[i].catfood;//此时准备的猫食以花干
printf("%.3lf\n",sum);
}
}
return 0;
}