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. 

InputThe 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. 

OutputFor 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

1 #include
     
       2 #include
      
        3 #include
       
         4 using namespace std; 5 struct kang 6 { 7     double a, b; 8 }; 9 bool cmp(const kang&a, const kang&b)10 {11     return 1.0*a.a / a.b > 1.0*b.a / b.b;12 }13 int main()14 {15     kang k[10000];16     int m, n;17     18     while (cin >> m >> n,n == -1 && m == -1)19     {20         double sum=0;21         for (int i = 0; i < n; i++)22             cin >> k[i].a >> k[i].b;23         sort(k, k + n , cmp);24         for (int i = 0; i < n&&m != 0; i++)25         {26             if (m - k[i].b >= 0)27             {28                 sum += k[i].a;29                 m -= k[i].b;    30             }31             else32             {33                 sum += m * 1.0*k[i].a / k[i].b;34                 m = 0;35             }36         }37         printf("%.3f\n", sum);38     }39     return 0;40 }
       
      
     

简单的贪心题，自定义一个cmp就看解决，不过在用sort时注意区间为左闭右开；

也可以使用符号重载：

bool operator < (const kang & a, const kang & b)

{

return a.a / a.b > b.a / b.b;

}（已测试）