number homework 1.5 gre with span 使用 lin amp

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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 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             else
32             {
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 }

簡單的貪心題，自定義一個cmp就看解決，不過在用sort時註意區間為左閉右開；

也可以使用符號重載：

bool operator < (const kang & a, const kang & b)
{
return a.a / a.b > b.a / b.b;
}（已測試）

FatMouse' Trade