A - FatMouse' Trade

最新推荐文章于 2021-03-21 17:27:11 发布

转载最新推荐文章于 2021-03-21 17:27:11 发布 · 150 阅读

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4248809.html

本文探讨了一种算法问题，即FatMouse如何通过最优策略利用有限的猫粮交换最大数量的JavaBeans。通过定义结构体存储房间信息，使用比较函数进行排序，并采用贪心算法策略，实现了在给定条件下获取最大利益的解决方案。

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A - FatMouse' Trade

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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 
     

       my answer: 
     

#include<iostream>
#include<algorithm>
using namespace std;
typedef struct room{
    double w;
    double v;
    double age;
}room;
bool cmp(room a,room b){
    if(a.age<b.age)return true;
    else return false;
}
int main()
{
    int m,n;
    while(cin>>m>>n)
    {
        if(m==-1&&n==-1)
            return 0;
        room a[1001];
        for(int i=0;i!=n;i++){
            scanf("%lf%lf",&a[i].w,&a[i].v);
            a[i].age=a[i].v/a[i].w;
        }
        sort(a,a+n,cmp);
        double sum=0.0;
        int i=0;
        while(m>0&&i<n)
        {
            if(m>=a[i].v){
                sum+=a[i].w;
                m-=a[i].v;
                i++;
            }
			else{
                sum+=(m*1.0/a[i].v)*a[i].w;
                m-=m;
                i++;
            }
        }
        printf("%.3f\n",sum);
    }
    return 0;
}

原来提交的时候超时，检查了一下，发现限制时间是10秒，我的事11秒多，于是把cin改成了scanf我的时间就缩短了近5秒，于是在提交，返回WA，想了想意识到可能存在M用不完的情况，再交就AC了

转载于:https://www.cnblogs.com/NYNU-ACM/p/4248809.html