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
题意:
有m的猫粮,n个房间,每个房间有一定量的javabean,在房间获取百分之多少的javabean,就要付出相应比例的猫粮,求m磅的猫粮最多可以获得多少javabean。
思路:
把每个房间可以获得的javabean和付出的猫粮比率存在一个数组,排序。从比率最大的房间开始,一直到没有猫粮了。
代码:
#include<stdio.h>
int main()
{
int n,m,i,j;
double sum,t;
double r[1010];
int je[1010],f[1010];
while(scanf("%d%d",&m,&n)!=EOF)
{
if(m==-1&&n==-1)
break;
sum=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&je[i],&f[i]);
r[i]=je[i]*1.0/f[i];
}
for(i=1;i<n;i++)
for(j=i+1;j<=n;j++)
{
if(r[i]<r[j])
{
t=r[i];
r[i]=r[j];
r[j]=t;
t=je[i];
je[i]=je[j];
je[j]=t;
t=f[i];
f[i]=f[j];
f[j]=t;
}
}
for(i=1;i<=n;i++)
{
if(m>f[i])
{
m-=f[i];
sum+=je[i];
}
else
{
sum+=r[i]*m;
break;
}
}
printf("%.3f\n",sum);
}
return 0;
}