HDU_1009——老鼠的交易,性价比排序,最大化收益

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

 1 /*

 2 算性价比j[i]/f[i],然后按性价比对每个房间的数据重新排序

 3 再累加最大化兑换比例的老鼠粮 

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 const int MAX = 1005;

 8 /*

 9 int compare(const void *a,const void *b)

10 {

11     //return *(double*)a - *(double*)b; //小——大 

12     return *(double*)b - *(double*)a;    //大——小 

13 }

14 */

15 int main()

16 {

17     int m,n,j[MAX]={0},f[MAX]={0};

18     while(~scanf("%d%d",&m,&n))

19         {

20             if(m==-1 && n==-1)

21                 break;

22             double j_f[MAX]={0},temp=0;

23             for(int i=1;i<=n;i++)

24                 {

25                     scanf("%d%d",&j[i],&f[i]);

26                     j_f[i]=(double)j[i]/f[i];

27                 }

28             //快速排序：数组首地址，元素个数，一个元素大小，指向比较函数的指针 

29             //qsort(j_f+1,n,sizeof(double),compare);

30             //好吧，写到一半发现排序不能这样用- -排序函数还是留着吧- -

31             for(int i=1;i<=n-1;i++)

32                 {

33                     for(int k=i+1;k<=n;k++)

34                         {

35                             if(j_f[i]<j_f[k])

36                                 {

37                                     temp=j_f[k];

38                                     j_f[k]=j_f[i];

39                                     j_f[i]=temp;

40                                     

41                                     temp=j[k];

42                                     j[k]=j[i];

43                                     j[i]=(int)temp;

44                                     

45                                     temp=f[k];

46                                     f[k]=f[i];

47                                     f[i]=(int)temp;

48                                 }

49                         }

50                 }

51             double ans=0;

52             for(int i=0;i<=n;i++)

53                 {

54                     if(m>=f[i])

55                         {

56                             ans+=j[i];

57                             m-=f[i];

58                         }

59                     else

60                         {

61                             ans+=m*j_f[i];

62                             break;

63                         }

64                 }

65             printf("%.3lf\n",ans);

66         }    

67     return 0;

68 }