ACM--猫鼠交易--HDOJ 1009--FatMouse' Trade--贪心

HDOJ题目地址：传送门

FatMouse' Trade

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

Total Submission(s): 67028    Accepted Submission(s): 22794

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

#include 
#include 
#include 
#include 
using namespace std;
struct node{
    int J;
    int F;
}Node[1100];
bool cmp(node a,node b){
    return 1.0*a.J/a.F>1.0*b.J/b.F;
}
int main(){
    int m,n;
    while(scanf("%d%d",&m,&n)&&m!=-1&&n!=-1){
        double result=0.0,java=m-0.0;
        int temp=0;
        while(tempNode[i].F){
                result+=Node[i].J;
                java-=Node[i].F;
            }else{
                result+=(Node[i].J*(java/Node[i].F));
                break;
            }
        }
       printf("%.3lf\n",result);
    }
}