HDOJ 1009之FatMouse' Trade

FatMouse' Trade

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

Total Submission(s): 66418    Accepted Submission(s): 22570

 

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 thatFatMouse 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	

 

大致题意：大肥鼠有m磅的猫粮，而大肥鼠最喜欢吃的JavaBean在仓库里，于是它要拿着猫粮去跟在仓库看守JavaBean的猫去换，仓库被分成n个部分。第i个部分的JavaBean有J[i]磅，需要F[i]磅猫粮兑换；大肥鼠并不用把该部分的全部JavaBean都兑换掉，而是按照每一部分的比例（权重）来兑换！你的任务就是算出大肥鼠能兑换JavaBean的最大值。

 

#include
#include 
using namespace std;
struct node{
    int j;
    int f;
    double w; //用w来表示各个部分的权重
}a[1001]; 
bool cmp(node a,node b){
	return a.w>b.w;//自定义比较函数
}
int main(){
	int m,n;
	while(scanf("%d %d",&m,&n)&&(m!=-1||n!=-1)){//输入格式要求，当m，n都为-1是终止
		for(int i=0;ia[i].f){
				sum+=a[i].j;//如果大肥鼠拥有的猫粮M大于兑换所用的猫粮，则直接全部兑换
				m-=a[i].f;
			}
			else{
				sum+=a[i].w*m;//否则M乘以比值
				break;
			}
		}
		printf("%.3lf\n",sum);//输出格式要求
	}
    return 0; 
}