贪心之FatMouse' Trade

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 
#include<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<deque>#include<list>using namespace std;struct wp{    double d,f,x;//float型数据在相除的时候会有偏差，用double型比较好，减少误差}p[1010];int cmp(wp a,wp b){    return a.x>b.x;}int main(){    int m,n,i;    double sum=0;    while((scanf("%d%d",&m,&n)!=EOF) &&(n!=-1 || m!=-1))    {        sum=0;        for(i=0;i<n;i++)        {            cin>>p[i].d>>p[i].f;            p[i].x=p[i].d/p[i].f;        }        sort(p,p+n,cmp);        for(i=0;i<n;i++)        {            if(p[i].f<m)            {                sum+=p[i].d;                m-=p[i].f;            }            else            {                sum+=p[i].x*m;                break;            }        }        printf("%.3f\n",sum);    }    return 0;}