HDU1009老鼠的旅行 (贪心算法)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62898    Accepted Submission(s):
21231

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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#include <string.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
struct sa
{
    int a,b;
    double c;
}data[];
bool cmp(sa x,sa y)
{
    return x.c>y.c;
}
int main()
{
    int i,j,m,n;
    double sum;
    while (cin>>m>>n&&m!=-&&n!=-)
    {
        sum=;
        for (i=;i<n;i++)
        {
            cin>>data[i].a>>data[i].b;
            data[i].c=(double)data[i].a/(double)data[i].b;
        }
        sort(data,data+n,cmp);
        //for (i=0;i<n;i++)
        //cout<<data[i].a<<" "<<data[i].b<<" "<<data[i].c<<endl;
        for (i=;i<n;i++)
        {
            if (m>=data[i].b)
            {
                m-=data[i].b;
                sum+=data[i].a;
            }
            else if (m<data[i].b&&m>)
            {
                sum+=data[i].c*(double)m;
                m=;
            }
            else
            break;
        }
        printf("%.3lf\n",sum);
    }
    return ;
}