二、stl ，模拟，贪心等 [Cloned] E - 贪心

原题：

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.

题意：

老鼠准备M磅的食物去找猫换豆子，猫有很多个仓库，每个仓库储存着数量不同的豆子，而且换取豆子的需要的食物都不相同，不过老鼠可以在每个仓库都换取一部分豆子，只需要付出相应比例的食物就可以。

题解：

简单的贪心问题，就是求出每个仓库的豆子/食物的比值作为价值，然后依次从价值大到价值小开始换取，到最后如果还有豆子但是不够换取下一个仓库的所有豆子，就用剩下的所有豆子换取这个仓库里的一部分豆子。

题解：AC

#include<stdio.h>
#include<stdlib.h>
struct rate
{
    double a;
    double b;
    double rt;
}rat[1010];
int cmp(const void *a, const void *b)
{
    return (*(rate *)b).rt > (*(rate *)a).rt ? 1:-1 ;
}
int main()
{
    int m, n, i;
    double sum, x;
    while(scanf("%d%d", &m, &n) != EOF &&( m != -1 || n != -1))
    {
        x = (double)m;
        for( i = 0; i  < n; i++)
        {
            scanf("%lf%lf", &rat[i].a, &rat[i].b);
            if( rat[i].b != 0)
            rat[i].rt = rat[i].a / rat[i].b;
            else
            rat[i].rt = 0;
        }
        qsort(rat, n, sizeof(rat[0]), cmp);
        sum = 0;
        for(i = 0 ; i < n; i++)
        {
            if( x >= rat[i].b )
            {
                sum += rat[i].a;
                x -= rat[i].b;
            }
            else
            {
                if(x > 0)
                {
                    sum += (x / rat[i].b) * rat[i].a;
                    x = 0;
                }
            }
        }
        printf("%.3lf\n", sum);
    }
    return 0;
}