FatMouse' Trade

 FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 

题目大意：一只老鼠要用猫粮和猫交易食物，有n个房间，给出每个房间里的食物和进入需要的猫粮，第一行两个数分别表示总共的猫粮数量m和房间数量n，接下来是n行是n个房间里的食物数量和需要猫粮数量，求老鼠能交换最多的食物数量。

/*
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。
*/

//AC
#include<stdio.h>
#include<algorithm>
#include<vector>
using namespace std;
struct data
{
double j,f,r;
};
bool cmp(struct data a,struct data b)
{
if (a.r>b.r)
return true;
return false;
}
int main()
{
int m,n;
while (scanf("%d%d",&m,&n))
{
if (m==-1&&n==-1)
break;
int i;
vector<struct data> D(1010);
for (i=0;i<n;i++)
{
scanf("%lf%lf",&D[i].j,&D[i].f);
D[i].r=D[i].j/D[i].f;
}
sort(D.begin(),D.end(),cmp);
double sum=0;
for (i=0;i<n;i++)
{
if (m>=D[i].f)
{

4000
sum+=D[i].j;
m-=D[i].f;
}
else
{
sum+=(D[i].j/D[i].f)*m;
break;
}
}
printf("%.3lf\n",sum);
}
return 0;
}