杭电1009（FatMouse' Trade）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51740    Accepted Submission(s): 17347

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

// 贪心，排序求最优解；

#include<stdio.h>
#include<algorithm> 
using namespace std;
struct d 
{
    double m,n;
    double value;
};

d sum[100000];

bool cmp(d m,d n)
{
    return m.value>n.value;
}

int main()
{
    int a,b;
    while(~scanf("%d %d",&a,&b))
    {
        if(a==-1&&b==-1)
        break;
        int i;
        double total=0,c=0;
        for(i=0;i<b;i++)
        {
            scanf("%lf %lf",&sum[i].m,&sum[i].n);
            sum[i].value=sum[i].m/sum[i].n;
        }
        
        sort(sum,sum+b,cmp);
        
    /*    for(i=0;i<b;i++) //为什么注释掉的部分for循环结束后，不加if语句OJ就一直错，搞不懂；
        {
            a-=sum[i].n;
            if(a>=0)
            total+=sum[i].m;
            else
            break;
        }*/
        for(i=0;i<b;i++)
        {
            if(a>sum[i].n)
            {
                total+=sum[i].m;
                a-=sum[i].n;
            }
            else
            {
                c=1;
                total+=a*sum[i].value;
            }
            if(c!=0)
            break;
        }
        /*if(a<0)
        total+=(a+sum[i].n)*sum[i].value; 
        */printf("%.3lf
",total);
    }
    return 0;
}