hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43036    Accepted Submission(s): 14376

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
/*题解：     简单的贪心问题，老方法，先排序     */#include<cstdio>#include<algorithm>using namespace std;struct FM{    double a;    double b;}e[10010];int cmp(FM a,FM b){    return a.a/(a.b*1.0)>b.a/(b.b*1.0);}int main(){    int m,n,i,j;    double sum,t;    while(scanf("%d %d",&m,&n)&&(n!=-1&&m!=-1))    {        for(i=0; i<n; i++)        scanf("%lf %lf",&e[i].a,&e[i].b);        sort(e,e+n,cmp);        for(i=0,sum=0.0,t=0.0; i<n; i++)        {            if(t+e[i].b<m)            {                sum+=e[i].a;                t+=e[i].b;            }            else            {                sum+=(m-t)*(e[i].a/(e[i].b*1.0));                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}