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
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<algorithm>
5 using namespace std;
6 int main()
7 {
8 int m,n;
9 int f[1200],j[1200];
10 double a[1200],b[1200];
11 int next[1000];
12 while((scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)))
13 {
14 for(int i=0;i<n;i++)
15 {
16 scanf("%d%d",&f[i],&j[i]);
17 a[i]=1.0*f[i]/j[i];
18 }
19 double num=0;
20 for(int i=0;i<n;i++)
21 {
22 for(int k=i+1;k<n;k++)
23 {
24 if(a[i]<a[k])
25 {
26 swap(a[i],a[k]);
27 swap(f[i],f[k]);
28 swap(j[i],j[k]);
29 }
30 }
31 if(j[i]<m)
32 {
33 num+=f[i];
34 m-=j[i];
35 }
36 else
37 {
38
39 num+=1.0*f[i]*m/j[i];
40 m=0;
41 }
42 if(m==0)
43 {
44 break;
45 }
46 }
47 printf("%.3lf\n",num);
48 }
49 }