HDU 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 80381    Accepted Submission(s): 27762

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：一只大老鼠有M磅猫粮，然后他准备去跟猫做个交易以取得他喜欢的javabean。猫管理的仓库有n个房间，每个房间有J磅javabean而且总共需要f磅猫粮来换，请问怎样换就实惠。

赤裸裸的贪心....算出每个房间的单价（表达有问题.....），然后就买呗

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct Node{double j,f,value;}cat[1010];bool cmp(Node a,Node b){return a.value>b.value;}int main(){int m,n;while(~scanf("%d%d",&m,&n)){if(m==-1&&n==-1) break;for(int i=0;i<n;i++){scanf("%lf%lf",&cat[i].j,&cat[i].f);cat[i].value=cat[i].j/cat[i].f;}sort(cat,cat+n,cmp);double sum=0;for(int i=0;i<n;i++){if(m>=cat[i].f){m-=cat[i].f;sum+=cat[i].j;}else{sum+=cat[i].value*m;break;}}printf("%.3lf\n",sum);}}