CDZSC_2015寒假新人（1）——基础 c

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<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 int main()
 {
     int m,n;
     int f[],j[];
     double a[],b[];
     int next[];
     while((scanf("%d%d",&m,&n)&&(m!=-||n!=-)))
     {
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&f[i],&j[i]);
             a[i]=1.0*f[i]/j[i];
         }
         double num=;
         for(int i=;i<n;i++)
         {
             for(int k=i+;k<n;k++)
             {
                 if(a[i]<a[k])
                 {
                     swap(a[i],a[k]);
                     swap(f[i],f[k]);
                     swap(j[i],j[k]);
                 }
             }
             if(j[i]<m)
             {
                 num+=f[i];
                 m-=j[i];
             }
             else
             {

                 num+=1.0*f[i]*m/j[i];
                 m=;
             }
             if(m==)
             {
                 break;
             }
         }
         printf("%.3lf\n",num);
     }
 }