1408111555-hd-FatMouse' Trade.cpp

                                     FatMouse' Trade

                                              Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                               Total Submission(s): 43146    Accepted Submission(s): 14413

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

 

题目大意

      老鼠拿着粮食去跟猫交换，老鼠有n粮食，有m个房间，每个房间的老鼠的粮食和猫需要的都不一样，但是都是等比例交换，也就是在每个房间老鼠拿着（猫所需粮）*a%可以换（房间内鼠粮）*a%，要求老鼠可以换到的最多的粮食数量。

解题思路

      很明显需要用到贪心，将兑换比例做降序排列，然后依次兑换直到老鼠手里没粮食为止。

代码

#include<stdio.h>#include<algorithm>using namespace std;struct food{    int j;    int f;    double h;}foods[1100];bool cmp(food a,food b){    return a.h>b.h;}int main(){    int n,m;    int i,k;    double sum;    while(scanf("%d%d",&n,&m),n!=-1||m!=-1)    {        for(i=0;i<m;i++)        {            scanf("%d%d",&foods[i].j,&foods[i].f);            foods[i].h=foods[i].j/(foods[i].f*1.0);        }        sort(foods,foods+m,cmp);        sum=0;        for(i=0;i<m;i++)        {            if(n>foods[i].f)            {                n-=foods[i].f;                sum+=foods[i].j;            }            else            {                sum+=n*foods[i].j/(foods[i].f*1.0);                n=0;            }            if(n==0)                break;        }        printf("%.3lf\n",sum);    }    return 0;}