FatMouse' Trade

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

Total Submission(s): 44470    Accepted Submission(s): 14872

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

题目大意：有N个房间，每一个房间存有FatMouse喜欢吃的食物。可是每一个房间

的食物都须要用对应的猫粮去换。

FatMouse 有M磅的猫粮，为它最多能换到多

少的食物。

思路：贪心方法。用结构体存每间房间的食物量和所需猫粮量。

按食物的单位价格（

即食物/猫粮的大小）进行排列，每次选单位价格最小的购买，知道M磅猫粮用完

#include
     
      #include
      
       #include
       
        using namespace std;struct warehouse{    double j;    double f;}a[1100];bool cmp(warehouse a,warehouse b){    return a.f/a.g < b.f/b.g;}int main(){    int N;    double M;    while(~scanf("%lf%d",&M,&N)&& (M!=-1||N!=-1))    {        memset(a,0,sizeof(a));        for(int i = 0; i < N; i++)        {            scanf("%lf%lf",&a[i].j,&a[i].f);        }        sort(a,a+N,cmp);        double sum = 0;        for(int i = 0; i < N; i++)        {            if(M <= 0.000001)                break;            if(M >= a[i].f)            {                sum += a[i].j;                M -= a[i].f;            }            else            {                sum += M*a[i].j/a[i].f;                M = 0;            }        }        printf("%.3lf\n",sum);    }    return 0;}