杭电oj 1009

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76080    Accepted Submission(s): 26061

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

刚写完的时候，测试数据都过了，但是提交就是不过，数据类型转换也没错啊，也都*1.0了，但是就是不对，最后把变量的类型都定义为double,结果就对了.....醉了。。

#include
#include
#include
#include
using namespace std;
struct node
{
    double j;   //之前定义为int
   double  f;    //之前定义为int
    double avg;
}s[1015];
int m,n;
double sum;
bool cmp(node a,node b)
{
    if(a.avg>b.avg)
        return true;
    else
        return false;
}
int main()
{


    while(scanf("%d%d",&m,&n)!=EOF)
    {
        sum=0;
        memset(s,0,sizeof(s));
        if(m==-1&&n==-1)
            break;
        else
        {
                for(int i=0;i=s[i].f)
                {
                    m=m-s[i].f;
                    sum=sum+s[i].j;
                }
                else
                {
                    sum=sum+s[i].j*(m/(s[i].f*1.0));
                    m=0;
                }


            }
                printf("%0.3lf\n",sum);
        }
        
        
    }
    return 0;
}