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.
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
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333 31.500
题意：给你m克猫粮，n个房间分别可以用猫粮以J[i]/F[i]的比例换取至多J[i]克咖啡豆，问最多能换取多少咖啡豆
分析：先排下序再贪心
#include <stdio.h>
#include <algorithm>
using namespace std;
int n,u;
double m,ans;
struct room
{
double j,f,p;
} r[];
int cmp(room a,room b)
{
return a.p>b.p;
}
int main()
{
while(scanf("%lf%d",&m,&n)&&n!=-&&m!=-)
{
for(int i=;i<n;i++)
scanf("%lf%lf",&r[i].j,&r[i].f),
r[i].p=r[i].j/r[i].f;
sort(r,r+n,cmp);
u=ans=;
while(m&&u<n)
{
if(r[u].f>m) ans+=m*r[u].p,m=;
//剩下的猫粮不够把这个房间的咖啡豆换过来，那就能换多少换多少
else ans+=r[u].j,m-=r[u].f;
//够的话，这间房子全部咖啡豆换过来
u++;
}
// while(m>=r[u].f&&u<n) //换种写法
// {
// ans+=r[u].j;
// m-=r[u].f;
// u++;
// }
// if(u!=n) ans+=m*r[u].p;
printf("%.3lf\n",ans); }
return ;
}