FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

这道题是道贪心的题目，就是题目有点难懂，很早之前就想做这题了，奈何题目一直看不懂，花了很大功夫理解了题目。然后今天终于做出来了。。题意：老鼠有m克的猫粮，想去跟猫兑换食物，但是老鼠喜欢的食物在仓库的不同层，每层的兑换率都不一样。求的是最多能兑换到多少食物。

5 3

7 2

4 3

5 2

这个案例代表有老鼠有5克猫粮 ，食物一共放在不同的三层 ，第一层有7克食物，需要2克猫粮兑换，兑换率为7/2；第二层第三层类似。

所以要贪心的话，就要先选出兑换率最大的一层开始兑换。上代码上解析。

#include <stdio.h>
#define wbx 50000
int a[wbx]; //wbx代表第wbx层，a[wbx]代表第wbx层的食物。
int b[wbx]; //b[wbx]第wbx层的猫粮。
double p[wbx];// p代表第wbx层的兑换率。
#include <string.h>
#include <algorithm>
using namespace std;
void solve()
{
	int m,n;
	while(scanf("%d%d",&m,&n)&&m!=-1 &&n!=1)
	{
		int i,k;
		double sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&a[i],&b[i]);//输入每层的食物和猫粮
			p[i]=double(a[i]*1.0/b[i]);  //算出每层的兑换率
		}
		while(m!=0 &&n!=0)  //m=0或者n=0代表着结束，跳出循环。
		{
			double max=-999;
			for(i=1;i<=n;i++)
			{
				if(p[i]>max)
				{
					max=p[i];//每次都先找出每层兑换率最大的。
					k=i;//把值赋给k。
				}
			}
			if(m-b[k]>=0)// 如果猫粮大于等于这层的猫粮
			{
				m-=b[k];//  减去这层的猫粮。
				sum+=a[k];// 得到这层所有食物。
				p[k]=0;// 这层兑换完了，让它的兑换率为0。
			}
			else
			{
				sum+=m*p[k];// 如果剩余猫粮已经小于想兑换的这层的猫粮，那就用剩下的猫粮兑换所有的食物。
				m=0;//猫粮为0，跳出循环。
			}
		}
		sum=double(sum);
		printf("%.3lf\n",sum);//sum就是最优解。。
	}
}
int main()
{
	solve();//问题解决。。
	return 0;
}