【贪心】【HDOJ-1009】FatMouse' Trade

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

 

/************************************************************************************************************
题意：老鼠用M猫粮换仓库里的鼠粮，仓库有n个房间，每个房间有j[i]鼠粮，需要f[i]猫粮换，
可以换部分猫粮
思路：简单的贪心问题，先算出性价比j[i]/f[i]，然后从大到小排序，先从性价比高的开始换
等到鼠粮或猫粮没有时结束
注意：此题有几个比较坑的数据  

0 1
1 0
1.000

1 0
0.000

5 4
10000 5
2000 2
100 0
300 0
10400.000
************************************************************************************************************/
#include <cstdio>
#include <algorithm>
using namespace std;
typedef struct 
{
    int j;
    int f;
    double va;
}node;
node a[1000+10];
bool cmp(node a, node b);
int main()
{
    int n , m;
    while(scanf("%d %d", &m , &n) && (m != -1 || n != -1))
    {
        double ans = 0;
        if(!n)
        {
            printf("%.3f
",ans);
            continue;
        }
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%d %d", &a[i].j, &a[i].f);
            if(!a[i].f)
            {
                ans += a[i].j;
                a[i].j = 0;
                continue;
            }
            a[i].va = a[i].j * 1.0 / a[i].f;
        }
        sort(a , a + n , cmp);
        for(int i = 0 ; m > 0 && i < n ; i++)
        {
            if(m >= a[i].f)
            {
                m -= a[i].f;
                ans += a[i].j;
            }
            else
            {
                ans += a[i].va * m;
                break;
            }
        }
        printf("%.3f
", ans);
    }
    return 0;
}

bool cmp(node a, node b)
{
    if(a.va > b.va)
        return true;
    return false;
}