hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43036    Accepted Submission(s): 14376


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
/*题解: 
    简单的贪心问题,老方法,先排序 
    */
#include<cstdio>
#include<algorithm>
using namespace std;
struct FM
{
    double a;
    double b;
}e[10010];
int cmp(FM a,FM b)
{
    return a.a/(a.b*1.0)>b.a/(b.b*1.0);
}
int main()
{
    int m,n,i,j;
    double sum,t;
    while(scanf("%d %d",&m,&n)&&(n!=-1&&m!=-1))
    {
        for(i=0; i<n; i++)
        scanf("%lf %lf",&e[i].a,&e[i].b);
        sort(e,e+n,cmp);
        for(i=0,sum=0.0,t=0.0; i<n; i++)
        {
            if(t+e[i].b<m)
            {
                sum+=e[i].a;
                t+=e[i].b;
            }
            else
            {
                sum+=(m-t)*(e[i].a/(e[i].b*1.0));
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}
   


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值