FatMouse' Trade（hdoj1009）

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

 #include<stdio.h>/**此题有很多边缘数据没有想到，或者压根就没有去想/
 #include<algorithm>
 using namespace std;
 typedef struct
 {
     double x,y;/*数据类型错点*/
     double z;
 }node;
 int cmp(node p,node q)
 {
     return p.z>=q.z;
 }
 node a[];
 int main()
 {
     double M;
     int N;
     while(~scanf("%lf%d",&M,&N))
     {
         int i,j;
         double max=,shen=M;
         if(M==-&&N==-)
             break;
         if(N==)/*room为0情况*/
             {
                 printf("0.000\n");
                 continue;
             }
         else
         {
             for(i=;i<N;i++)
                 {
                     scanf("%lf%lf",&a[i].x,&a[i].y);
                     a[i].z=a[i].x/a[i].y;
                 }
             sort(a,a+N,cmp);
             /*for(i=0;i<N;i++)
                 printf("%d %d  %lf\n",a[i].x,a[i].y,a[i].z);*/
             for(i=;;i++)
             {
                 if(shen>a[i].y)
                 {
                     max+=a[i].x;
                     shen=shen-a[i].y;
                 }
                 else if(shen==a[i].y)/*可能漏掉，因为存在y=0情况,此情况不能和else并在一起，看了讨论组才知道的*/
                 {
                     max+=a[i].x;
                     shen=shen-a[i].y;
                 }
                 else
                 {
                     max+=(shen/a[i].y)*a[i].x;
                     break;
                 }
             }
         }
         printf("%.3lf\n",max);
     }
 }