1009：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

 

 

代码实现：

#include<cstdio>

double java[100000],f[100000],a[100000];

int main(){ double m; int n; scanf("%lf%d",&m,&n); while(m!=EOF&&n!=EOF) {  double res=0;  for(int i=0;i<n;i++)  {   scanf("%lf%lf",&java[i],&f[i]);   a[i]=java[i]/f[i];  }  for(i=0;i<n;i++)  {     int maxnum=0;   for(int j=1;j<n;j++)   {    if(a[maxnum]<a[j])     maxnum=j;   }   if(m>f[maxnum])   {    res+=java[maxnum];    m-=f[maxnum];    a[maxnum]=-1;   }   else   {    res+=m*a[maxnum];    break;   }  }  printf("%.3lf/n",res);  scanf("%lf%d",&m,&n); } return 0;}