杭电ACM1009

题目：

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.

（这里没用百度翻译或者其他翻译是因为不太好翻译，大体的意思就是有只老鼠想和猫做交易，用猫粮换JavaBeans，老鼠有数量为m的猫粮，有n间房子里面有j[i]的JavaBeans，但是需要f[i]的猫粮交换）

样例输入：

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

样例输出

13.333
31.500

思路：
这个就是一个典型的贪心算法，用结构体存储数据j[i]和f[i];再开一个double 类记录j[i]/f[i]的大小；用sort排序结构体就行；

上代码

#include<iostream>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int book[100005];

#define ll long long

struct node{
	int JavaBeans;
	int catfood;
    double s;
	
}a[100005];
int cmp(node a,node b)
{
  if(a.s>b.s)
  return 1;
  else
  return 0;	
	
}
int main()
{   
  int i,j,n,m;
  
  while(scanf("%d %d",&m,&n)!=EOF)
  {
  	if(n==-1||m==-1)
  	break;
  	for(i=1;i<=n;i++)
  	{
  	  cin>>a[i].JavaBeans>>a[i].catfood;
  	  a[i].s=(double)(a[i].JavaBeans*1.0)/(a[i].catfood *1.0);
  		
	}
  	sort(a+1,a+1+n,cmp);
  	int sum=0;
  	double sum1=0;
  	for(i=1;i<=n;i++)
  	{
  	  sum1+=a[i].JavaBeans;
	  sum+=a[i].catfood;
	  if(sum>=m)
	  {
	  	sum1-=a[i].JavaBeans;
	  	sum1+=(m-(sum-a[i].catfood))*a[i].s;
	  	break;
	  }
  		
	}
	printf("%.3lf\n",sum1);
  	
  } 
    
 return 0; 
}