FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57207 Accepted Submission(s): 19174
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct zz
{
int v;
int w;
double c;
}q[1010];
int cmp(zz a,zz b)
{
return a.c>b.c;
}
int main()
{
int n,m,i,j;
while(scanf("%d%d",&m,&n),n!=-1&&m!=-1)
{
for(i=1;i<=n;i++)
{
scanf("%d%d",&q[i].v,&q[i].w);
q[i].c=q[i].v*1.0/q[i].w;
}
sort(q+1,q+n+1,cmp);
double sum=0;
for(i=1;i<=n;i++)
{
if(q[i].w<=m)
{
sum+=q[i].v;
m-=q[i].w;
}
else
{
sum+=m*q[i].c;
break;
}
}
printf("%.3lf\n",sum);
}
return 0;
}
