杭电-FatMouse' Trade

╰+哭是因爲堅強的太久メ 2022-08-04 17:59 126阅读 0赞

# FatMouse' Trade #

##### Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) #####

##### Total Submission(s) : 59   Accepted Submission(s) : 17 #####

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### Problem 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

###     ###

### AC代码如下 ###

#include <iostream>
    #include <iomanip>
    using namespace std;
    class Trade
    {
    public:
        int javab;
        int food;
        double bili()
        {
            return 1.0*javab/food;
        }
    };
    int main()
    {
        int m,n;
        Trade trade[1000];
        while(cin>>m>>n&&m!=-1&&n!=-1)
        {
            for(int i=0; i<n; ++i)
                cin>>trade[i].javab>>trade[i].food;
            int i,k,j;
            Trade t;
            for(i=0; i<n-1; i++)
            {
                k=i;
                for(j=i+1; j<n; j++)
                {
                    if(trade[k].bili()<trade[j].bili())
                        k=j;
                }
                t=trade[k];
                trade[k]=trade[i];
                trade[i]=t;
            }
            double sum=0;
            for(int i=0; i<n&&m!=0; ++i)
            {
                if(m>=trade[i].food)
                {
                    m-=trade[i].food;
                    sum+=trade[i].javab;
                }
                else
                {
                    sum+=m*(trade[i].bili());
                    m=0;
                }
            }
            cout<<fixed<<setprecision(3)<<sum<<endl;
        }
        return 0;
    }

心得：

吐血了。。。这英文我也是醉了，一边翻译一边错

用GUN C++是报错的，用VS C++才AC，尼玛。。。。。。。。。。

心得：

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[Center]: /images/20220805/e1535240e8614d94a4e79e7e92d24096.png