航电ac 1009 FatMouse' Trade

航电ac 1009 FatMouse' Trade
FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12577 Accepted Submission(s): 3707
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：
#include
main(){
int n,i,k;
float j[1001],f[1001],s[1001],t1,t2,m;
double max;
while(scanf("%f%d",&m,&n)&&(n!=-1||m!=-1)){
max=0;
for(i=0;i

#include
using namespace std;
int k[1001],f[1001];
double ma[1001];
int main()
{double m,n;
while(cin>>m>>n,m!=-1,n!=-1)
{for(int i=1;i>k[i]>>f[i];
ma[i]=double(k[i])/f[i];
}
for(i=1;i