1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45213    Accepted Submission(s): 15137

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

 

Author

CHEN, Yue

 

 

Source

ZJCPC2004

 

 1 #include<stdio.h> 2 #include<math.h> 3 #include<string.h> 4 #include<stdlib.h> 5 struct ln{ 6     double x; 7     double y; 8     double weight; 9 }a[1005];10 int cmp(const void*a,const void*b)11 {12     return (*(struct ln*)a).weight<(*(struct ln*)b).weight?1:-1;13 }14 int main()15 {16     //freopen("in.txt","r",stdin);17     int m,n;18     while(~scanf("%d%d",&m,&n))19     {20         memset(a,0,sizeof(a));21         if(m==-1&&n==-1)22         break;23         for(int i=0;i<n;i++)24         {25             scanf("%lf%lf",&a[i].x,&a[i].y);26             a[i].weight=a[i].x/a[i].y;27         }28         qsort(a,n,sizeof(struct ln),cmp);29         double sum=0;30 31         for(int i=0;i<n;i++)32         {33             if(m>=a[i].y)34             {35                 sum+=a[i].x;36                 m-=a[i].y;37             }38             else39             {40                 sum+=a[i].weight*m;41                 m=0;42             }43         }44         printf("%.3lf\n",sum);45     }46     return 0;47 }

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