FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45213 Accepted Submission(s): 15137
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.
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
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
struct ln{
double x;
double y;
double weight;
}a[];
int cmp(const void*a,const void*b)
{
return (*(struct ln*)a).weight<(*(struct ln*)b).weight?:-;
}
int main()
{
//freopen("in.txt","r",stdin);
int m,n;
while(~scanf("%d%d",&m,&n))
{
memset(a,,sizeof(a));
if(m==-&&n==-)
break;
for(int i=;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
a[i].weight=a[i].x/a[i].y;
}
qsort(a,n,sizeof(struct ln),cmp);
double sum=; for(int i=;i<n;i++)
{
if(m>=a[i].y)
{
sum+=a[i].x;
m-=a[i].y;
}
else
{
sum+=a[i].weight*m;
m=;
}
}
printf("%.3lf\n",sum);
}
return ;
}