HDU1009FatMouse' Trade

FatMouse' Trade
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 3 Accepted Submission(s) : 3
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

Sample Output


13.333
31.500

Author
CHEN, Yue

Source
ZJCPC2004

贪心算法HDU1009 FatMouse' Trade

题目大意：

老鼠有M磅猫食。有N个房间，每个房间前有一只猫，房间里有老鼠最喜欢的食品JavaBean,J[i]。若要引开猫，必须付出相应的猫食F[i]。当然这只老鼠没必要每次都付出所有的F[i]。若它付出F[i]的a%，则得到J[i]的a%。求老鼠能吃到的做多的JavaBean。

解题思路：

老鼠要获得最多的食品，就要用最小的猫食换取最多的猫食，这就要求J[i]/F[i]的比例要大。J[i]/F[i]的比例越大，证明在这个房间，小鼠付出得到的收获最有价值。于是我们将设置结构体，结构体里设置percent放置J[i]/F[i]。然后对结构体数组进行排序。依次按比例排序的付出猫食，即可。

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
struct node
{
    double f;
    double j;
    double s;
};
bool cmp(node a,node b)
{
    return a.s>b.s;
}
int main()
{
    double m;
    int n;
    double sum;
    node nodes[1005];
    while(~scanf("%lf%d",&m,&n))
    {
        if(m==-1&&n==-1)
        break;
        sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&nodes[i].j,&nodes[i].f);
            nodes[i].s=(double)nodes[i].j/nodes[i].f;
        }
        sort(nodes,nodes+n,cmp);
        for(int k=0;k<n;k++)
        {
            if(m>=nodes[k].f)
            {
               sum=sum+nodes[k].j;
               m=m-nodes[k].f;
            }
            else
            {
                sum=sum+m*nodes[k].s;
                m=0;
                break;
            }
        }
        printf("%.3lf
",sum);
    }
}