CDZSC_2015寒假新人（1）——基础 c

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 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 #include<algorithm>

 5 using namespace std;

 6 int main()

 7 {

 8     int m,n;

 9     int f[1200],j[1200];

10     double a[1200],b[1200];

11     int next[1000];

12     while((scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)))

13     {

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

15         {

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

17             a[i]=1.0*f[i]/j[i];

18         }

19         double num=0;

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

21         {

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

23             {

24                 if(a[i]<a[k])

25                 {

26                     swap(a[i],a[k]);

27                     swap(f[i],f[k]);

28                     swap(j[i],j[k]);

29                 }

30             }

31             if(j[i]<m)

32             {

33                 num+=f[i];

34                 m-=j[i];

35             }

36             else

37             {

38                 

39                 num+=1.0*f[i]*m/j[i];

40                 m=0;

41             }

42             if(m==0)

43             {

44                 break;

45             }

46         }

47         printf("%.3lf\n",num);

48     }

49 }

View Code