【结构体排序】HDU1009FatMouse' Trade

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

题目意思：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。

英文题真的是很让人头痛啊！！！

WA了我好几次！！不能用C++的控制精度来输入输出！！！很可能会导致答案的精度出错！！

代码：

#include<iostream>#include<algorithm>#include<iomanip>#include<cstdio>using namespace std;struct node{    double c,v,price;}room[1010];bool cmp(node a,node b){    if(a.price!=b.price) return a.price>b.price;    else return a.v>b.v;}int main(){    int n,m;    //cin.sync_with_stdio(false);    //while(cin>>m>>n&&(n!=-1&&m!=-1)){    while(scanf("%d%d",&m,&n)!=EOF&&(n!=-1&&m!=-1)){        for(int i=0;i<n;i++){            scanf("%lf%lf",&room[i].v,&room[i].c);            //cin>>room[i].v>>room[i].c;            room[i].price=room[i].v/room[i].c;        }        sort(room,room+n,cmp);        double ans=0;        for(int i=0;i<n;i++){            if(m>=room[i].c){                ans+=room[i].v;                m-=room[i].c;            }else{                ans+=m*1.0*room[i].price;                break;            }        }        printf("%.3lf\n",ans);        //cout<<fixed<<setprecision(3)<<ans<<endl;      //  会产生精度问题；WA！！！！    }    return 0;}