Hdu ::FatMouse' Trade

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

贪心

#include <stdio.h> #include <algorithm>#include <iostream >using namespace std;  #define MAX 1005 struct node{ int J,K; double P; }house[MAX];  bool cmp(node a,node b) { return a.P>b.P; } int main(){ int n,m; while (scanf("%d%d",&m,&n),~m||~n)     {  for (int i=0;i<n;i++){ cin>>house[i].J>>house[i].K;        house[i].P=(double)house[i].J/house[i].K; } sort(house,house+n,cmp);  double sum=0; for (int i=0;i<n&&m;i++) {   if (m>=house[i].K){ m-=house[i].K;  sum+=house[i].J;  }else{ sum+=house[i].J*((double)m/house[i].K); m=0; } }printf("%.3lf\n",sum);   }      return 0; }