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): 77129    Accepted Submission(s): 26485

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

 

思路  : 按照  交换性价比来排序:  当大于时 按 100%   小于时  性价比乘以 剩下数量

#include <iostream>#include <stdio.h>#include <algorithm>#include <string>#include <cstring>#include <cmath>#include <queue>using namespace std;typedef long long ll;struct node{    double f,j,p;}s[10010];int cmp(node a,node b){    return a.p>b.p;}int main(){    int i,m,n;    while(~scanf("%d %d",&m,&n))    {        if(m==-1&&n==-1)            break;        double sum=0,res=m;        for(i=0;i<n;i++)        {            cin>>s[i].j>>s[i].f;            s[i].p=s[i].j/s[i].f;        }        sort(s,s+n,cmp);//        for(i=0;i<n;i++)//        {//            printf("%lf %lf %lf\n",s[i].j,s[i].f,s[i].p);//        }        for(i=0;i<n;i++)        {            if(res>s[i].f)            {                sum+=s[i].j;                res-=s[i].f;            }            else            {                sum+=res*s[i].p;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}

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