hdu 1009 FatMouse' Trade（贪心）

 

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22453    Accepted Submission(s): 6992

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

 

贪心：按J[i]/F[i]降序，依次找最大的就行了

#include<iostream>
#include<algorithm>
#include<cstdio>
#define N 1010
using namespace std;

struct SS{double r,j,f;}a[N];

bool cmp(SS A,SS B)
{
    if(A.r>B.r) return true;
    return false;
}

int main()
{
    int n,m,i;
    double ans;
    while(cin>>m>>n&&(n!=-1&&m!=-1))
    {
        for(i=0;i<n;i++) 
        {
            cin>>a[i].j>>a[i].f;
            a[i].r=(a[i].j/a[i].f);
        }
        sort(a,a+n,cmp);
        ans=0;
        for(i=0;i<n;i++)
        {
            if(a[i].f<=m) 
            {
                ans+=(a[i].j);
                m-=a[i].f;
            }
            else 
            {
                ans+=a[i].r*m;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}