杭电 1009 FatMouse' Trade （贪心）

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

 

 

 

题目大意：

老鼠有M磅猫食。有N个房间，每个房间前有一只猫，房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫，必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%，则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 struct stu
 {
     double a,b;  //不用float
 }a[];
 bool cmp(stu a,stu b)
 {
     return (a.a/a.b)>(b.a/b.b);
 }
 int main()
 {
     int m,n,i,j;
     double sum;
     while(scanf("%d %d",&m,&n)&&(m!=-)&&(n!=-))
     {
         sum=;
         for(i=; i < n; i++)
         {
             scanf("%lf %lf",&a[i].a,&a[i].b);
         }
         sort(a,a+n,cmp);
         for(i=;i<n;i++)
         {
             if(m > a[i].b)
             {
                 sum+=a[i].a;
                 m-=a[i].b;
             }
             else
             {
                 sum+=((m/a[i].b)*a[i].a);
                 break;
             }
         }
         printf("%.3lf\n",sum);
     }
 }