杭电ACM 第1009题 FatMouse' Trade

杭电ACM 第1009题 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
我的代码如下：
#include
#define M 100
void main()
{
\x05int m,n,i,k,j[M],f[M],t1,t2;
\x05double a[M],temp,s;
\x05while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))
\x05{
\x05\x05i=0;
\x05\x05k=n;
\x05\x05s=0;
\x05\x05while(k--)
\x05\x05{
\x05\x05\x05scanf("%d%d",&j[i],&f[i]);
\x05\x05\x05a[i]=(double)j[i]/f[i];
\x05\x05\x05i++;
\x05\x05}
\x05\x05for(i=0;i