FatMouse' Trade

 FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 37 24 35 220 325 1824 1515 10-1 -1 

 

Sample Output

 13.33331.500 

 

#include<iostream>using namespace std;struct room{double catfood;double mousefood;double aver;};int main(){  int M,N,i,j,k;  while(cin>>M>>N)  {  room a[1000];  double sum=0;  if(M==-1&&N==-1)  break;  else  {     for(i=0;i<N;i++)  {  cin>>a[i].mousefood>>a[i].catfood;  }  for(i=0;i<N;i++)  {  a[i].aver=a[i].mousefood/a[i].catfood;//相同猫食(消耗)时候可以得到的鼠食(利益)  }         for(i=0;i<N-1;i++) { j=i; for(k=j+1;k<N;k++) { if(a[k].aver>a[j].aver) { j=k; } } if(i!=j) { room t;   //相同消耗，得到的最多 的降序排列 t=a[j]; a[j]=a[i]; a[i]=t; } } for(i=0;i<N;i++) { if(a[i].catfood<M) {                 sum=sum+a[i].mousefood; M=M-a[i].catfood; }              else  {  sum=sum+M*(a[i].mousefood)/a[i].catfood;  break;  } } printf("%.3lf\n",sum);   }  }  return 0;}