FatMouse＆＃39; Trade（杭电1009）

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FatMouse’ Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 1

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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

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Sample Output

13.333
31.500

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//本题是简单贪心问题。意思是老鼠有m榜猫粮，通过给猫粮东西data[i].b能够兑换data[i].a，求最多能够兑换多少食物。//仅仅需将data[i].a/data[i].b的值从大到小排序就可以求解。 #include<stdio.h>struct st{    double a;    double b;    double c;}data[1000];int main(){    int i,j,m,n;    struct st data[1000],t;    while(scanf("%d %d",&m,&n)&&(m!=-1||n!=-1))    {        double sum=0.000;        for(i=0;i<n;i++)        {            scanf("%lf %lf",&data[i].a,&data[i].b);        }        for(i=0;i<n;i++)        {            for( j=i+1;j<n;j++)            {                if((data[i].a/data[i].b)<data[j].a/data[j].b)                {                t=data[i];                data[i]=data[j];                data[j]=t;}            }        }        for(i=0;i<n;i++)        {            if(m-data[i].b>=0.001)            {                sum+=data[i].a;                m-=data[i].b;            }            else            {                sum=sum+m*data[i].a/data[i].b;                break;            }        }         printf("%.3lf\n",sum);    }    return 0;}

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