Hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43156    Accepted Submission(s): 14417

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

 

Recommend

JGShining

 

 

贪心.

贪心策略是根据价值比进行降序排序 ：J[i]/F[i];

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define MAX 1001
struct node
{
    int f;
    int j;
    double p;
};
struct node x[MAX];
int n,m;
bool cmp(struct node x,struct node y)
{
    if(x.p==y.p) return x.f<y.f;
    return x.p>y.p;
}
void out()
{
    for(int i=0;i<n;i++)
        cout<<x[i].p<<" ";
    cout<<endl;
}
void init()
{
    memset(x,0,sizeof(x));
}
void read()
{
    int i;
    for(i=0;i<n;i++)
    {
        scanf("%d %d",&x[i].j,&x[i].f);
        x[i].p=(double)(x[i].j*1.0/(x[i].f*1.0));
    }
    sort(x,x+n,cmp);
}
void cal()
{
    int i;
    double res=m;
    double ans=0;
    for(i=0;i<n;i++)
    {
        if(res>=x[i].f)
        {
            res-=x[i].f;
            ans+=x[i].j;
        }
        else
        {
            ans=ans+res*x[i].p;
            res-=res;
        }
    }
    printf("%.3f
",ans);
}
void solve()
{
    init();
    read();
    cal();
}

int main()
{
    while(scanf("%d %d",&m,&n)!=EOF)
    {
        if(n==-1&&m==-1) break;
        solve();
    }
    return 0;
}