HDU 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50683 Accepted Submission(s): 16959


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

题意：

一共有n个食物，有m种豆子，接下来m行表示的是 k个食物能交换每种的j个豆子，最多能换多少个豆子。

#include <stdio.h>struct food{    double F;    double J;    double ave;}a[1001];int main (void){    //freopen("1009.txt","r",stdin);    double n;    int i,j,k,m;    struct food t;    double sum;    while(scanf("%lf%d",&n,&m)==2&&n!=-1&&m!=-1)    {        sum=0;        for(i=1;i<=m;i++)        {            scanf("%lf%lf",&a[i].F,&a[i].J);            a[i].ave=a[i].F/a[i].J;        }        for(i=1;i<m;i++)        {            k=i;            for(j=i+1;j<=m;j++)            {                if(a[k].ave<a[j].ave)                {                    k=j;                }            }                    t=a[i];                    a[i]=a[k];                    a[k]=t;        }        for(i=1;i<=m;i++)        {            if(n>=a[i].J)            {                 n-=a[i].J;                 sum+=1.0*a[i].F;            }            else            {                sum+=n*a[i].ave;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}