1433：FatMouse Trade

• 题目描述：

  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.

• 输入：

  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.

• 输出：

  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.

• 样例输入：

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

• 样例输出：

  13.333
  31.500

利用贪心法，我们先计算各种物品的性价比，依次买性价比最高的物品，直到钱花完为止。

AC代码为：

    #include <iostream>
    #include <stdio.h>
    #include <algorithm>

    using namespace std;

    struct goods{ //表示可买物品的结构体
        double j; //物品总重
        double f; //物品总价值
        double s; //物品总性价比
        bool operator < (const goods &A) const{
            return s>A.s;
    }
    }buf[1000];

    int main()
    {
        double m;
        int n;
        while(scanf("%lf%d",&m,&n)!=EOF){
            if(m==-1 && n==-1)
                break;
            for(int i=0;i<n;i++){
                scanf("%lf%lf",&buf[i].j,&buf[i].f); //输入
                buf[i].s=buf[i].j/buf[i].f; //计算性价比
            }
            sort(buf,buf+n);
            int idx=0;
            double ans=0;
            while(m>0 && idx<n){ //既有钱剩余和物品剩余时继续循环
                if(m>buf[idx].f){
                    ans+=buf[idx].j;
                    m-=buf[idx].f;
                }else{
                    ans+=buf[idx].j*m/buf[idx].f;
                    m=0;
                }
                idx++;
            }
            printf("%.3lf\n",ans);
        }
        return 0;
    }
    /*
    5 3
    7 2
    4 3
    5 2
    20 3
    25 18
    24 15
    15 10
    -1 -1
    */

这是王道书上的一道简单例题，详细的贪心算法的具体内容还有待我后续去学习！