FatMouse’ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45918    Accepted Submission(s): 15370

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

#include <cstdio>
#include <algorithm>
using namespace std;
struct trade
{
	double a;
	double b;
	double c;
}fj[1000];//WA了一次，错误是运行中错误，所以最有可能是数组开的太小了，干脆把数组长度从100开到1000，就AC了！
int compare(trade a,trade b)
{
	return a.c>b.c;//运用结构体变量存储，
}
int main()
{
    int m,n,i;
	double j,f,sum;
	while (scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1))
	{
		sum=0;
        for (i=0;i<n;i++)
        {
			scanf("%lf%lf",&j,&f);
			fj[i].a=j;
			fj[i].b=f;
			fj[i].c=j/f;
        }
		sort(fj,fj+n,compare);//用结构体中的c变量的大小进行排序，注意这个是从大到小进行！
		for (int j=0;j<n;j++)
		{
			if (m-fj[j].b>0.001)
			{
				sum+=fj[j].a;
				m-=fj[j].b;
			}//当m大于整体的b时
			else
			{
				sum+=m*fj[j].a/fj[j].b;
				break;
			}//当m不能整体换时！
		}
		printf("%.3lf\n",sum);
	}
}//这一题典型的