FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 45970 Accepted Submission(s): 15397
Problem DescriptionFatMouse 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.InputThe 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. OutputFor 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 Input5 37 24 35 220 325 1824 15 15 10-1 -1Sample Output
#include<stdio.h>
void bubblesort(double v[],int w[],int f[],int n)
{
int i,j;
double t;
for(i=;i<=n;i++)
{
for(j=i+;j<=n;j++)
{
if(v[i]>v[j])
{
t=v[i];
v[i]=v[j];
v[j]=t; t=f[i];
f[i]=f[j];
f[j]=t; t=w[i];
w[i]=w[j];
w[j]=t;
}
}
}
}
int main()
{
int n,m,i,w[],f[];
double v[],sum;
while(scanf("%d %d",&m,&n)!=EOF&&(n!=-)&&(m!=-))
{
for(i=;i<=n;i++)
{
scanf("%d %d",&w[i],&f[i]);
v[i]=f[i]*1.0/w[i];
}
bubblesort(v,w,f,n); sum=;
for(i=;i<=n&&v[i]<=m&&m>;i++)
{ if(m>=f[i])
{
sum+=w[i];
m=m-f[i];
}
else
{
sum+=m*1.0/v[i];
m=;
}
}
printf("%.3lf\n",sum);
}
}