C - FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45965    Accepted Submission(s): 15395

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

 

 

首先这是一道贪心算法，当f/j越大时代表这样越有利，所以进行排序；

但也有值得注意的几组数据（来自杭电的大神）

此题除了要满足例子以外，还要满足一些条件才能真正算ac：
0 1
1 0
1.000

1 0
0.000

5 4
10000 5
2000 2
100 0
300 0
10400.000

数据类型用double,就这样

#include<stdio.h>
int j[1010],f[1010];
double a[1010];
void paixu(double a[],int k[],int f[],int n)
{
    int i=0,j=n-1,h=k[0],m=f[0];
    double  t=a[0];
    if(n>1){
        while(i<j)
        {
            for(;i<j;j--)
            if(a[j]>t){
                a[i]=a[j];
                f[i]=f[j];
                k[i]=k[j];
                i++;
                break;
            }
            for(;i<j;i++)
            {
                if(a[i]<t){
                    a[j]=a[i];
                    f[j]=f[i];
                    k[j]=k[i];
                    j--;
                    break;
                }
            }
        }
        a[i]=t;
        f[i]=m;
        k[i]=h;
        paixu(a,k,f,i);
        paixu(a+i+1,k+i+1,f+i+1,n-i-1);
    }
}
void hanshu(int m,int n)
{
    int i;
    double  sum=0;
    for(i=0;i<n;i++)
    {
        if(m>=f[i]){
            sum=sum+j[i];
            m=m-f[i];
        }
        else{
            sum=sum+1.0*m/f[i]*j[i];
            break;
        }
    }
    printf("%0.3f\n",sum);
}
int main()
{
    int n,m,i,t;
    while(1)
    {scanf("%d%d",&m,&n);
    if(m==-1&&n==-1)break;
    for(i=0;i<n;i++)
        {scanf("%d%d",&j[i],&f[i]);
        if(f[i]!=0)a[i]=1.0*j[i]/f[i];
        else a[i]=1020;
        }
        paixu(a,j,f,n);
        hanshu(m,n);
        }
    return 0;

}