FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 14

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

Sample Output

13.333
31.500

Author

CHEN, Yue

Source

很简单的贪心题，

先按照价值/代价的比值来排序，肯定是先买比值大的。

 

#include<iostream>
#include<algorithm>
using namespace std;
struct stu
{
	int a;
	int b;
	double rate;
}s[1005];
bool cmp(stu x,stu y)
{   
	if(x.rate!=y.rate)
	return x.rate>y.rate;
    else   
	return x.a<y.a;
}
int main()
{   //freopen("1.txt","r",stdin);
	int m,n,i;
	double cnt;
	while(scanf("%d %d",&m,&n)!=EOF)
	{   if(m==-1&&n==-1)break;
		cnt=0;
		for(i=0;i<n;i++)
		{
			scanf("%d %d",&s[i].a,&s[i].b);
			s[i].rate=s[i].a*1.0/s[i].b;
		}
		sort(s,s+n,cmp);
		for(i=0;i<n;i++)
		{
			if(m>=s[i].b)
			{
				m-=s[i].b;
				cnt+=s[i].a;
			}
           else
		   {
			   cnt+=s[i].rate*m;break;
		   }
		}
     printf("%.3lf\n",cnt);
	}
	return 0;
}