FatMouse' Trade http://acm.hdu.edu.cn/showproblem.php?pid=1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34126    Accepted Submission(s): 11152

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

 

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JGShining

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef struct sss
{
	double x;
	double y;
	double z;
}sss;
struct sss a[1010];
int cmp(const void *a,const void *b)
{
	sss *c=(sss *)a;
	sss *d=(sss *)b;
	return c->z>d->z?1:-1;

}

int main()
{
	int m,n;
	while(scanf("%d %d",&m,&n),(n!=-1||m!=-1))
	{
		int i;
		double toal=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf %lf",&a[i].x,&a[i].y);
			a[i].z=a[i].x/a[i].y;
		}
		qsort(a,n,sizeof(a[0]),cmp);
		for(i=n-1;i>=0;i--)
		{

			if(m<=a[i].y)
			{
				toal+=a[i].x*m/a[i].y;
				break;
			}
			else
			{
				toal+=a[i].x;
				m-=a[i].y;
			}
		}
		printf("%.3lf
",toal);
	}
	return 0;
}

先理解一下题意：老鼠和猫交易，规则很简单，每个仓库里有J[i]磅鼠食，如果全部换走这些，老鼠需要F[i]磅猫食，当然老鼠也不用把一个仓库里的鼠食全部换走，可以按比例来换，比如

5 3

7 2

4 3

5 2

意思就是老鼠有5磅猫食，先用2磅猫食换7磅鼠食（7 2），然后还剩3磅猫食，再用2磅猫食换5磅鼠食（5 2），还剩1磅，然后换1.33磅（4*1/3），所以一共最多换13.333磅鼠食。

理解了意思就好做了，一看就是贪心算法，肯定要排序。排序时要注意，输入的J[i]F[i]一定用double不然的话计算结果double z=J[i]/f[i]是整型，我的就是那种情况，这和书上讲的不太一样，好像用int型时它只是把结果转换成了double型。