FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 37292 Accepted Submission(s): 12311



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

ZJCPC2004

include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
struct info
{
int j;
int f;
double ave;
}unit[10005];
int cmd(const info &x,const info &y)
{
//printf("fhdshfjks");
return x.ave<y.ave;
}
int main()
{
int i,j,m,n;
double sum;
//info unit[10005];
while(scanf("%d%d",&m,&n)!=EOF)
{
if((n==-1)&&(m==-1))
break;
for(i=0;i<n;i++)
{
scanf("%d%d",&unit[i].j,&unit[i].f);
unit[i].ave=unit[i].f/(unit[i].j*1.0);
}
sort(unit,unit+n,cmd);
//for(i=0;i<n;i++)
//printf("%f\n",unit[i].ave);
sum=0;
//printf("%d",m);
for(i=0;i<n&&m!=0;i++)
{
if(m>=unit[i].f)
{
sum=sum+unit[i].j;
//printf("hdjks%f\n",sum);
m=m-unit[i].f;
}
else
{
sum=sum+(m/(unit[i].f*1.0))*unit[i].j;
//printf("aaa%f\n",sum);
m=0;
break;
}

}

printf("%.3f\n",sum);
}
return 0;

}

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