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.
InputThe 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.
OutputFor 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 -1Sample Output
13.333
31.500
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
struct node
{
double bean,food;
}s[1010];
bool cmp(node a,node b)
{
if(a.bean/a.food>b.bean/b.food)return true;
return false;
}
int main()
{
double m;
int i,n;
while(scanf("%lf%d",&m,&n)){
if(m==-1&&n==-1)break;
for(i=0;i<=n-1;i++){
scanf("%lf%lf",&s[i].bean,&s[i].food);
}
sort(s,s+n,cmp);
double ans=0;
for(i=0;i<=n-1;i++){
if(m>=s[i].food){
m-=s[i].food;
ans+=s[i].bean;
}else
{
ans+=m/(s[i].food)*s[i].bean;
m=0;
break;
}
}
printf("%.3lf\n",ans);
}
return 0;
}
先买bean/food大的
