Hdu1009_fatmouse 'trade [greedy] [water question]

Fatmouse 'trade

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/others) total submission (s): 44470 accepted submission (s): 14872
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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1-1
 
Sample output
13.333
31.500
 
Author
Chen, Yue
 
Source

Zjcpc2004

There are n rooms, and each room contains the food that fatmouse prefers, but each room

All the food needs to be replaced with the appropriate cat food. Fatmouse has m-pound cat food, so it can be changed to a maximum

Less food.

Idea: greedy method. Use a struct to store the amount of food in each room and the amount of cat food required. Per unit price of food (

That is, the size of food/cat food), the minimum unit price of each purchase, know m pound cat food used up

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct warehouse{    double j;    double f;}a[1100];bool cmp(warehouse a,warehouse b){    return a.f/a.g < b.f/b.g;}int main(){    int N;    double M;    while(~scanf("%lf%d",&M,&N)&& (M!=-1||N!=-1))    {        memset(a,0,sizeof(a));        for(int i = 0; i < N; i++)        {            scanf("%lf%lf",&a[i].j,&a[i].f);        }        sort(a,a+N,cmp);        double sum = 0;        for(int i = 0; i < N; i++)        {            if(M <= 0.000001)                break;            if(M >= a[i].f)            {                sum += a[i].j;                M -= a[i].f;            }            else            {                sum += M*a[i].j/a[i].f;                M = 0;            }        }        printf("%.3lf\n",sum);    }    return 0;}

Hdu1009_fatmouse 'trade [greedy] [water question]