FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49098 Accepted Submission(s): 16485
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.
简单贪心....
需要注意的是数据是非负，所以有0的情况要考虑周全，基本都要特殊处理。
多WA了三次，不知道为什么交C++可以过，交G++就不行。
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <iomanip>
using namespace std;
int main()
{
int m,n,f[1500],j[1500];
double scale[1500];
double ans,sum;
while(scanf("%d %d",&m,&n)!=EOF&&(m!=-1))
{ ans=0;
// if (m==0)
memset(scale,0,sizeof(scale));
memset(f,0,sizeof(f));
memset(j,0,sizeof(j));
// bool flag=false;
for (int i=1;i<=n;i++)
{
scanf("%d %d",&j[i],&f[i]);
if (f[i]!=0)
scale[i]=(double)j[i]*1.0/f[i];
else if (j[i]!=0)
{
ans=ans+j[i];
}
}
// if (flag) {cout<<fixed<<setprecision(3)<<ans<<endl;continue;}
for (int i=1;i<=n-1;i++)
for (int k=i+1;k<=n;k++)
if (scale[i]<scale[k])
{
swap(scale[i],scale[k]);
swap(j[i],j[k]);
swap(f[i],f[k]);
}
sum=0;
int i=1;
while (m>=sum&&i<=n)
{
ans=ans+j[i];
sum=sum+f[i];
i++;
}
i--;
ans=ans-j[i];
sum=sum-f[i];
ans=ans+(m-sum)*scale[i];
cout<<fixed<<setprecision(3)<<ans<<endl;
}
return 0;
}