HDOJ--1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58243    Accepted Submission(s): 19502

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

胖老鼠...出题人很有想法，典型的贪心算法再加上一个快排即可AC。既然想获得最多食物，那么获得的与支出的比值越大则越划算。

以下是我的AC代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;struct node{    double x,y;}a[1000];bool cmp(node a1,node a2){    double b1,b2;    b1=double(a1.x/a1.y);    b2=double(a2.x/a2.y);    return b1>b2;}int main(){        int m,n;    while(scanf("%d%d",&m,&n)==2)    {        if (m==-1 && n==-1)break;        double max=0;        for(int i=0;i<n;i++)        {            cin>>a[i].x>>a[i].y;        }        sort(a,a+n,cmp);        for(int i=0;i<n;i++)        {            if(m>=a[i].y)            {                m-=a[i].y;                max+=a[i].x;            }            else            {                max+=a[i].x*m/a[i].y;                m=0;            }            if(m==0)break;        }        cout.precision(3);        cout<<fixed<<max<<endl;    }    return 0;}