HDU 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 59851    Accepted Submission(s): 20095

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

 

 

使用格式输入输出的格式说明符%d,%f,%lf等，应当注意与int，float，double等类型的对应关系。

 

 

 

#include <cstdio>
#include <algorithm>
using namespace std;

const int MAX=1050;
struct Room{
    double j;
    double f;
    double w;
}r[MAX];

bool compare(const Room &r1,const Room &r2)
{
    if(r1.w>r2.w)
        return true;
    else
        return false;
}

int main()
{
    double m,ans,jv,fv;
    int n;
    while(scanf("%lf%d",&m,&n)==2&&n!=-1)
    {
        ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&jv,&fv);
            r[i].j=jv;
            r[i].f=fv;
            r[i].w=jv/fv;
        }
        sort(r,r+n,compare);

        for(int i=0;i<n;i++)
        {
            if(m>=r[i].f)
            {
                m-=r[i].f;
                ans+=r[i].j;
            
            }
            else
            {
                ans+=r[i].w*m;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
}