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  • HDU 1009:FatMouse' Trade

    FatMouse' Trade

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 60163    Accepted Submission(s): 20217


    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
     

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    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    struct Node
    {
        double j,f,p;
    } node[10000];
    int cmp(Node x,Node y)
    {
        return x.p>y.p;
    }
    int main()
    {
        int m,n;
        while(~scanf("%d%d",&n,&m) && (m!=-1 || n!=-1))
        {
            double sum = 0;
            int i;
            for(i = 0; i<m; i++)
            {
                scanf("%lf%lf",&node[i].j,&node[i].f);
                node[i].p = node[i].j/node[i].f;
            }
            sort(node,node+m,cmp);
            for(i = 0; i<m; i++)
            {
                if(n>node[i].f)
                {
                    sum+=node[i].j;
                    n-=node[i].f;
                }
                else
                {
                    sum+=node[i].p*n;
                    break;
                }
            }
            printf("%.3lf
    ",sum);
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/im0qianqian/p/5989592.html
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