FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37292    Accepted Submission(s): 12311

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

Author

CHEN, Yue

 

Source

ZJCPC2004

include<cstdio>#include<algorithm>#include<iostream>using namespace std;struct info{    int j;    int f;    double ave;}unit[10005];int cmd(const info &x,const info &y){    //printf("fhdshfjks");    return x.ave<y.ave;}int main(){    int i,j,m,n;    double sum;    //info unit[10005];    while(scanf("%d%d",&m,&n)!=EOF)    {     if((n==-1)&&(m==-1))     break;     for(i=0;i<n;i++)     {         scanf("%d%d",&unit[i].j,&unit[i].f);         unit[i].ave=unit[i].f/(unit[i].j*1.0);     }     sort(unit,unit+n,cmd);     //for(i=0;i<n;i++)     //printf("%f\n",unit[i].ave);     sum=0;     //printf("%d",m);     for(i=0;i<n&&m!=0;i++)     {         if(m>=unit[i].f)         {             sum=sum+unit[i].j;             //printf("hdjks%f\n",sum);             m=m-unit[i].f;         }         else         {             sum=sum+(m/(unit[i].f*1.0))*unit[i].j;             //printf("aaa%f\n",sum);             m=0;             break;         }     }     printf("%.3f\n",sum);    }    return 0;}

 

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