ZOJ Problem Set - 2109 FatMouse' Trade

ZOJ Problem Set - 2109

FatMouse' Trade

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Time Limit: 2 Seconds                                     Memory Limit: 65536 KB                            

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

Sample Output

  13.333
  31.500

明显的贪心算法

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct St{
	int j,f;
	double jpf;
}; 
bool cmp(St a,St b)
{
	return a.jpf>b.jpf;
}
int main()
{
	int m,n;
	while(cin>>m>>n)
	{
		if(m==-1&&n==-1)break;
		St s[1001];
		for(int i=0;i<n;i++)
		{
			 cin>>s[i].j>>s[i].f;
			 s[i].jpf=s[i].j/(double)s[i].f; 
		}
		sort(s,s+n,cmp);
		double sum=0.0; 
		for(int i=0;i<n;i++)
		{
			if(m>s[i].f)
			{
			   sum+=s[i].j;
			   m-=s[i].f;
			}
			else if(m>0)
			{
				sum+=s[i].j*(m/(double)s[i].f);
				m=0;
				break;
			}
		}
		printf("%.3f\n",sum);
	}
}