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): 66479    Accepted Submission(s): 22595

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题目描述：一共有n个房子，每个房子里有老鼠喜欢吃的javabeans,但是每个房间里的javabeans的价格不一样。老鼠用m元，问m元最多可以卖多少javabeans，其中每个房间里的javabeans可以被分割。
#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;struct luo{    double x,y,z;}lu[10000];bool cmp(luo a,luo b){    return a.z>b.z;}int main(){    int n,m;    while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1)    {double sum=0;        for(int i=0;i<n;i++)        {scanf("%lf%lf",&lu[i].x,&lu[i].y);        lu[i].z=lu[i].x/lu[i].y;｝//换算成单价        sort(lu,lu+n,cmp);        for(int i=0;i<n;i++)        {            if(m>=lu[i].y)            {sum+=lu[i].x;            m=m-lu[i].y;}            else            {                sum+=(m*lu[i].z);                break;            }        }        printf("%.3lf\n",sum);    }}