hdu 1009 FatMouse' Trade

FatMouse' Trade

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

Total Submission(s): 53352    Accepted Submission(s): 17788

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

有m个猫食n个房间，每一个房间都有一个猫把守。每一个房间都有j个你想要的东西，要得到东西。就要给猫猫食
	每一个猫都有规定的猫食，你给的猫食所占规定的比例就是你得到东西的比例，球最多能得多少个
简单贪心。排序，求值
2015,7,20


#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node{
	int j,f;
	double v;
}a[1100];
bool cmp(node x,node y){
	return x.v<y.v;
}
int main(){
	int m,n,i;
	while(~scanf("%d%d",&m,&n),!(m==-1&&n==-1)){
		for(i=0;i<n;i++){
			scanf("%d%d",&a[i].j,&a[i].f);
			a[i].v=a[i].f*1.0/a[i].j;
		}
		sort(a,a+n,cmp);
		double sum=0;
		for(i=0;i<n;i++){
			if(m>=a[i].f){
				sum+=a[i].j;
				m-=a[i].f;
			}
			else{
				double c;
				c=m*1.0/a[i].f;
				sum+=a[i].j*c;
				break;
			}
		}
		printf("%.3lf\n",sum);
	}
	return 0;
}