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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

#include
   
    #include
    
     #include
     
      using namespace std;     struct cat_mouse {         double mouse_food;         double cat_food;         double value;     } ; bool cmp(cat_mouse a,cat_mouse b) {         return a.value>b.value;} int main() {        double m,s;        int n;        struct cat_mouse a[1000];        while((cin>>m>>n)&&!(m==-1&&n==-1)) {             s=0;            for(int i=0;i< n;i++) {        cin>>a[i].mouse_food>>a[i].cat_food;                a[i].value=1.0*a[i].mouse_food/a[i].cat_food;            }          sort(a ,a+n,cmp);            for(int i=0;i
      
       =a[i].cat_food)                 {                  m-=a[i].cat_food;                 s+=a[i].mouse_food;              }                else{                    s+=1.0*m*a[i].mouse_food/a[i].cat_food; break;                     }                  }                 cout<
       
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c++ 编译通过了 但是G++ 没通过； 这是一个简单的贪心算法；

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