FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37520 Accepted Submission(s): 12395
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.
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
这是一道关于贪心的问题。这道题的大意是就是给定猫食的总量然后去和猫换JAVABEAN，每个房间都有JAVABEAN，当然每个房间需要的猫食不同。你要求的是利用有限的猫食换取最多的JAVABEAN。
#include<stdio.h>
int main()
{
int temp1,k,s,i,food,home;
while(scanf("%d%d",&food,&home)!=EOF&&(food!=-1&&home!=-1))
{
double a[1001];
int f[1001];
int j[1001];
double sum=0;
double temp;
for(i=0;i<home;i++)
{
scanf("%d%d",&j[i],&f[i]);
a[i]=j[i]*1.0/f[i]*1.0;
}
for(k=0;k<home;k++)
for(s=0;s<home-1;s++)
{
if(a[s]<a[s+1])
{
temp=a[s];
a[s]=a[s+1];
a[s+1]=temp;
temp1=f[s];
f[s]=f[s+1];
f[s+1]=temp1;
temp1=j[s];
j[s]=j[s+1];
j[s+1]=temp1;
}
}
for(i=0;i<home;i++)
{
if(food>=f[i])
{
sum=sum+j[i];
food=food-f[i];
}
else
{
sum=sum+food*a[i];
break ;
}
}
printf("%.3f\n",sum);
}
return 0;
}