HDOJ 1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54869    Accepted Submission(s): 18403

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
  
  
  
  

题意：老鼠一共有m斤猫食，xxx一共有n个仓库。每个仓库有a b两种数据，a表示当前仓库一共有多少斤鼠食，b表示当前仓库要换a斤鼠食要多少猫食。求老鼠最多能换多少食物。（http://blog.csdn.net/acmdream/article/details/18138335）

这道题要注意两点：

1.对其兑换比率进行排序；

2.记得在循环判断中先判断是否跳出，否则出错。

#include<stdio.h>
#include<algorithm>
using namespace std;

struct node{
	double a, b, percent;
};

node food[1010];

int cmp(node a, node b){
	return a.percent > b.percent;
}

int main(){
	int n, m;
	while(~scanf("%d%d", &n, &m), n != -1){
		int i;
		for(i = 0; i < m; ++i){
			scanf("%lf%lf", &food[i].a, &food[i].b);
			food[i].percent = food[i].a / food[i].b;
		}
		sort(food, food + m, cmp);
		double sum = 0.0;
		for(i = 0; i < m; ++i){
			if(n < food[i].b){
				sum = sum + n * food[i].percent;
				break;
			}
			if(n >= food[i].b){
				sum += food[i].a;
				n -= food[i].b;
			}
		}
		printf("%.3lf\n", sum);
	}
	return 0;
}