FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 41006    Accepted Submission(s): 13575

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

【题目来源】

【题目大意】

jack有M磅猫食，他想用这些猫食来换他最喜爱的javabeans，在他面前有n个房间，每个房间上表明了：J颗豆可以用F磅猫食来换取。

让你选择最优的方案换取最多的豆。

【题目分析】

贪心的水题，先排序，然后就按顺序选，直至将所有的猫食都换光。

 

#include
     
      #include
      
       using namespace std;struct Node{    int a,b;    double c;};Node node[1010];bool cmp(Node a,Node b){    return a.c>b.c;}int main(){    int n,m;    while(scanf("%d%d",&n,&m),n!=-1&&m!=-1)    {        for(int i=0;i
       
        node[i].b)                {                    sum+=node[i].a;                    n-=node[i].b;                }                else                {                    sum+=n*node[i].c;                    break;                }            }        }        printf("%.3lf\n",sum);    }    return 0;}