hdu-1009~(贪心)

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

FatMouse准备了M磅的猫食，准备与守卫仓库的猫交易，这些猫包含他最喜欢的食物，JavaBean。
仓库有N个房间。第i间房间包含J [I]磅的JavaBeans，并且需要F [i]磅的猫粮。 FatMouse不必交易房间内的所有JavaBeans，相反，如果他付给F [i] * 1磅的猫粮，他可能会得到1磅的JavaBeans。这里是一个实数。现在他正在为你分配这个作业：告诉他他可以获得的最大JavaBeans数量。


输入

输入由多个测试用例组成。每个测试用例都以包含两个非负整数M和N的行开始。然后N行包含两个非负整数J [i]和F [i]。最后一个测试用例后面跟着两个-1。所有整数不大于1000。


产量

对于每个测试用例，在一行中打印一个真实数字，精确到3位小数，这是FatMouse可以获得的最大JavaBean数量。


示例输入

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1


示例输出

13.333
31.500

解题思路：贪心思想，算出单价排序，然后从大到小去取.看代码

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <algorithm>

using namespace std;
struct stu
{
    double ja,mao,dj;
} c[1000];
bool cmp(stu x,stu y)
{
    return x.dj>y.dj;
}
int main()
{
    int m,n;
    while(scanf("%d %d",&m,&n),m!=-1&&n!=-1)
    {
        double a[5000],b[5000],ans=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%lf %lf",&c[i].ja,&c[i].mao);
            c[i].dj=c[i].ja/c[i].mao;
        }
        sort(c+1,c+n+1,cmp);
       for(int i=1;i<=n;i++)
        {
            if(m>=c[i].mao)
            {
                m-=c[i].mao;
                ans+=c[i].ja;
            }
            else
            {
                ans+=m*c[i].dj;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}