FatMouse' Trade

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
   
   
   
   

 

 

Author

CHEN, Yue

 

 

Source

ZJCPC2004

 

 

Recommend

JGShining

答案：

#include <stdio.h>

#define len 1000

template <class T>
void swap(T& x, T& y)
{
 T tmp = x;
 x = y;
 y = tmp;
}

int main(int argc, char* argv[])
{
 int m, n, i, j;
 int J[len], F[len];
 double a[len], v;

 while (scanf("%d %d", &m, &n)!=EOF)
 {
  if (m == -1 && n == -1)
   break;
  else if (n == 0)
  {
   printf("%.3f/n", 0);
   continue;
  }
  else if (m < 0 || n < 0)
   continue;

  for (i = 0; i < n; ++i)
   scanf("%d %d", &J[i], &F[i]);
  for (i = 0; i < n; ++i)
   a[i] = (double) J[i]*1.0/(F[i]*1.0);
  for (i = 0; i < n-1; ++i)
   for (j = i+1; j < n; ++j)
   {
    if (a[i] < a[j])
    {
     swap(a[i], a[j]);
     swap(J[i], J[j]);
     swap(F[i], F[j]);
    }
   }
  v = 0;
  for (i = 0; i < n; ++i)
  {
   if (m < F[i])
   {
    v += a[i]*m;
    break;
   }
   else
   {
    v += J[i];
    m = m - F[i];
   }
  }
  printf("%.3f/n", v);
 }

 return 0;
}