Fatmouse

最新推荐文章于 2023-05-09 13:40:37 发布

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本文介绍了一道关于猫鼠交易的编程题目，老鼠通过不同数量的猫粮换取仓库中的JavaBeans，目标是最大化获取的JavaBeans总量。文章提供了两种解决方案的代码示例，对比了它们之间的差异，并解释了为何一种方案能正确解决问题而另一种则不行。

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题目：

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

一个被自己蠢到笑死的题。。。很正常的想到了性价比，也想到了0 1和1 0 0 1的特殊情况，猫粮剩下的情况，然后光荣的wa。。。一大早起来看，原来是觉得没什么关系的循环中的判断段顺序造成的。
下面是过了的代码：

#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<set>
using namespace std;
struct foo
{
    double a, b;
    double ave;
}f[1005];

bool cmp(foo c,foo d)
{
    return c.ave>d.ave;
}
int main()
{
    double m;
    int n;
    while(cin>>m>>n)
    {
        if(m<0&&n<0) break;

        else
        {
            for(int i=0;i<n;i++)
        {
            cin>>f[i].a>>f[i].b;
            f[i].ave=f[i].a/f[i].b*1.0;//

        }
        sort(f,f+n,cmp);
        double ans=0;
        int i=0;
        while(m>=0)
        {
           if(i==n) break;
           if(m<=f[i].b)
            {
                ans+=m*f[i].ave;
                break;
            }
                else if(m>f[i].b)
            {
                m-=f[i].b;
                ans+=f[i].a;
                i++;
            }
        }
            printf("%.3lf\n",ans);
            }
    }
    return 0;
}

wa的代码：

#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<set>
using namespace std;
struct foo
{
    double a, b;
}f[1005];

bool cmp(foo c,foo d)
{
    return c.a/c.b>d.a/d.b;//最好两个数直接比较吧
}
int main()
{
    double m;
    int n;
    while(cin>>m>>n)
    {
        if(m<0&&n<0) break;
        if(n==0||m==0) //首先是这里，没房间没猫粮，哪会有豆子可换，直接就输出0，但是两个问题，1.没猫粮有房间，说明还有输入，2.真有不需要猫粮也可以换豆子的房间的话，n/0=INF，和0没关系啊。真的蠢到了，而且事实证明，并不需要单独拎出来这两种情况。

        {
            int u=0;
            cout<<u;
            continue;
        }
        else
        {
            for(int i=0;i<n;i++)
        {
            cin>>f[i].a>>f[i].b;
        }
        sort(f,f+n,cmp);
        double ans=0;
        int i=0;
        while(m>0)//一定要先判断是否m<=f[i].b,一个例子是0 1 1 0
        {
            if(i==n) break;
            if(m>=f[i].b)
            {
                m-=f[i].b;
                ans+=f[i].a;
                i++;
            }
            else if(m<f[i].b)
            {
                ans+=m*f[i].a/f[i].b;
                i++;
            }
        }
            printf("%.3lf\n",ans);
        }


    }
    return 0;
}