标签：clu += work i++ inpu roo val Requires bottom
FatMouse‘ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 95182 Accepted Submission(s):
33160
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
#include<stdio.h>
#define max 10000
int J[max],F[max];
double value[max];
int main()
{
int n,m;
while(scanf("%d%d",&m,&n))
{
if(n==-1&&m==-1) break;
double ans=0;
for(int i=0;i<n;i++)
{
scanf("%d%d",J+i,F+i);
value[i]=1.0*J[i]/F[i];
}
for(int i=0;i<n-1;i++)
for(int j=i+1;j<n;j++)
if(value[i]<value[j])
{
int t;
t=J[i];J[i]=J[j];J[j]=t;
t=F[i];F[i]=F[j];F[j]=t;
double t1;
t1=value[i];value[i]=value[j];value[j]=t1;
}
int i=0;
while(m>0&&i<n)//注意，这里i<n不能掉，因为M很可能大于需求量
{
if(m-F[i]>=0)
ans+=J[i];
else
ans+=1.0*m/F[i]*J[i];
m-=F[i++];
}
printf("%.3lf\n",ans);
}
return 0;
}
HDU 1009 FatMouse' Trade
标签：clu += work i++ inpu roo val Requires bottom
原文地址：https://www.cnblogs.com/shenyuling/p/9692469.html
