《ACM程序设计》书中题目P(会贿赂猫的胖老鼠)

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

题目大意就是一只胖老鼠会用猫粮和猫换食物，每间房的猫“汇率”都不一样，问如何换才能得到最多的食物；

思路如下：

1.根据每间房需要的猫粮和能换的食物来进行计算，算出“汇率”；

2.对汇率进行排序，选出交换的优先级；

3.一个一个计算直到猫粮耗尽为止。

程序如下：

#include
using namespace std;
struct Food
        {
                double j;
                double f;
                double a;
        };
bool cmp(const Food &a, const Food &b)
{
     return a.a>b.a;
}
int main()
{
        Food x[1000];
        int m,n,i;
        double s;
        while(cin>>m>>n)
        {
                if(m==-1)
                        break;
                s=0;
                for(i=0;i>x[i].j>>x[i].f;
                        x[i].a=x[i].j/x[i].f;  //计算每间房的汇率
                }
                sort(x,x+i,cmp);   //对汇率进行排序
                for(i=0;i=x[i].f)
                        {
                                m-=x[i].f;
                                s+=x[i].j;
                        }
                        else
                        {
                                s+=(x[i].a*m);
                                break;
                        }
                }
                cout<

要注意保留3位小数。