FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47476    Accepted Submission(s): 16005

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

#include<iostream>#include<algorithm>using namespace std;struct room{    int catfood;    int mousefood;    double danwei;};int cmp(room a,room b){       return a.danwei>b.danwei;}int main(){    int M,N,i;    room a[1000];    while(scanf("%d%d",&M,&N))    {        double sum=0;        if(M==-1 && N==-1)            break;        for(i=0;i<N;i++)        {            cin>>a[i].mousefood>>a[i].catfood;        }        for(i=0;i<N;i++)           a[i].danwei=a[i].mousefood*1.0/a[i].catfood;//单位猫食，获得的物品量        sort(a,a+N,cmp);//降序排列        i=0;        while(i<N && a[i].catfood<=M)        {            sum=sum+a[i].mousefood;            M=M-a[i].catfood;            i++;        }        if(i==N)        {            printf("%.3lf\n",sum);//准备的猫食可能有剩余        }        else        {            sum=sum+a[i].mousefood*M*1.0/a[i].catfood;//此时准备的猫食以花干            printf("%.3lf\n",sum);        }        }    return 0;}