FatMouse' Trade

题目：

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

这个问题和背包问题几乎一样，只是有点细微差别。

要注意的是，有可能某个房间里有JavaBean但是需要的猫粮是零，这种情况要考虑。

代码：

#include<stdio.h>
main()
{
int i,j;
int a[1100],b[1100],tempx;
double sum;
double c[1100],temp;
int m,n;
while(scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1))
{
sum=0;
for(i=0;i<n;i++)
{
scanf("%d%d",&a[i],&b[i]);
}
for(i=0;i<n;i++)
{
if(b[i]==0)
{
sum=sum+a[i];
for(j=i;j<n-1;j++)
{
a[j]=a[j+1];
b[j]=b[j+1];

}
n--;
i--;
}
}

for(i=0;i<n;i++)
{
c[i]=1.0*a[i]/b[i];
}
for(i=0;i<n;i++)
{
for(j=0;j<n-1;j++)
{
if(c[j]<c[j+1])
{
temp=c[j];
c[j]=c[j+1];
c[j+1]=temp;
tempx=a[j];
a[j]=a[j+1];
a[j+1]=tempx;
tempx=b[j];
b[j]=b[j+1];
b[j+1]=tempx;
}
}
}

for(i=0;i<n;i++)
{
if(m>b[i])
{
sum=sum+a[i];
m=m-b[i];
}
else
{
sum=sum+m*c[i];
break;
}

}

printf("%.3lf\n",sum);
}
}

还有同学的一个：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
double a;
double b;
double c;
}s[1000];
bool cmp(node a,node b)
{
return a.c>b.c;
}
int main()
{
double p,q;
int i,m,n;
while(scanf("%d%d",&m,&n),(m!=-1)&&(n!=-1))
{
p=0;
q=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&s[i].a,&s[i].b);
s[i].c=s[i].a/s[i].b;
}
sort(s,s+n,cmp);
i=0;
while(p<m&&i<n)
{
p+=s[i].b;
q+=s[i].a;
i++;
}
printf("%.3f\n",q-(p-m)*(s[i-1].a/s[i-1].b));

}
}