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.

该题目为贪心算法，将比例按顺序排好，按取所需就可！

 

 

#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;

struct Trade
{
 int j,f;
 double percent;
}Info[1005];

bool cmp(Trade &a,Trade &b)
{
 return a.percent>b.percent;
}

int main()
{
 int m,n;
 while(cin>>m>>n)
 {
  if(m==-1&&n==-1)
   break;
  for(int i=0;i<n;i++)
  {
   cin>>Info[i].j>>Info[i].f;
   Info[i].percent = (double)Info[i].j/Info[i].f;
  }
  double beans = 0,catfood = 0;
  int i;
  sort(Info,Info+n,cmp);
  for(i=0;i<n;i++)
  {
   beans += Info[i].j;
   catfood += Info[i].f;
   if(catfood>m)
   {
    catfood -= Info[i].f;
    beans -= Info[i].j; 
    break;
   }
  }
  if (i<n) 
   beans += (double)(m - catfood)/Info[i].f * Info[i].j;
  cout.precision(3);
  cout.setf(ios::fixed);
  cout<<beans<<endl;
 }
 return 1;
}