hdu1009 FatMouse' Trade(贪心)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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Total Submission(s): 51313    Accepted Submission(s): 17182

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

13.333 31.500

Author

CHEN, Yue

Source

​​ZJCPC2004​​

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解析：贪心。

           按照每个房间的 j[i]/f[i] 的值，有大到小进行排序，从最大的开始，能卖多少就买多少，直到买完为止。

           需要注意的是，有的房间的 f[i] 为 0 的情况，就是不要猫粮，直接白送（我靠，真是一只好猫，跟老鼠做交易就算了，还白送，我要是主人的话，我保证绝对不打死它）。

代码：

#include<cstdio>
#include<algorithm>
#define maxn 1000
using namespace std;

struct tnode{int x,y;double z;};
tnode a[maxn+20];

bool cmp(tnode p,tnode q)
{
  return p.z>q.z;
}

int main()
{
  freopen("1.in","r",stdin);
  int m,n,i,j,k,sum; double ans;
  while(scanf("%d%d",&m,&n)&&m!=-1)
    {
      ans=0,sum=0;
      for(i=1;i<=n;i++)
        {
          scanf("%d%d",&j,&k);
          if(k==0){ans+=j;continue;}
          a[++sum].x=j,a[sum].y=k;
          a[sum].z=j*1.0/k;
        }
      sort(a+1,a+sum+1,cmp);  
      for(i=1;m>=0 && i<=sum;i++)
        {
          k=min(m,a[i].y);
          ans+=a[i].z*k;
          m-=k;
        }
      printf("%.3lf\n",ans);  
    }
  return 0;
}