Hdoj 1009 FatMouse：【题解】

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 (分配) 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

• 理解一下题目：这个人准备了M pounds的food，准备用food和JavaBean兑换。要保证将M的food全部兑换，问最多能得到多少JavaBeans。在兑换时每个房间都存在一种比率，看你如何兑换？
• 注意本题有浮点数以及浮点数之间的运算，所以应该将除N之外的数都定义为double类型。
• 要点：贪心算法、sort对结构体的排序、sort对数组排序前要满足数组下标从0开始（不一定，当从n开始时需要在第一、二参数后加n）、牢记sort的头文件有 algorithm。

AC代如下码：

#include <iostream>
#include <algorithm>
using namespace std;
#define N 1005
struct warehouse{
	double j, f, price;
};
warehouse ware[N];
bool cmp(warehouse a, warehouse b){
	return a.price >= b.price;
}
int main(){
	int n, i;
	double max, m;
	while(scanf("%lf%d",&m,&n) != EOF){
		if(m == -1 && n == -1)
			break;
		for(i=0;i<n;i++){
			scanf("%lf%lf",&ware[i].j,&ware[i].f);
			ware[i].price = ware[i].j / ware[i].f;
		}
		sort(ware, ware+n, cmp);
		max = 0;
		for(i=0;i<n&&m;i++){
			if(m >= ware[i].f){
				m = m-ware[i].f;
				max+=ware[i].j;
			}
			else{
				max+=m/ware[i].f*ware[i].j;
				m = 0;
			}
		}
		printf("%.3lf\n",max);
	}
	return 0;
}