1020 moon cake

1020 The moon cake

Moon cake is a kind of traditional food that Chinese people eat during the Mid Autumn Festival , There are many different flavors of moon cakes in different regions . Now, the inventory of all kinds of mooncakes is given 、 Total selling price 、 And the biggest demand in the market , Please calculate the maximum profit you can get .

Be careful ： Allow to take out part of the inventory when selling . The example shows this ： If we have 3 Mooncakes , The inventory is 18、15、10 Ten thousand tons of , The total selling price is 75、72、45 One hundred million yuan . If the market's maximum demand is only 20 Ten thousand tons of , Then our biggest revenue strategy should be to sell all 15 Ten thousand tons 2 Mooncakes 、 as well as 5 Ten thousand tons 3 Mooncakes , get 72 + 45/2 = 94.5（ One hundred million yuan ）.

Input format ：

Each input contains a test case . Each test case shall be provided with no more than one 1000 The positive integer  N  Number of moon cakes 、 And not more than 500（ In 10000 tons ） The positive integer  D  Represents the maximum market demand . The next line shows  N  A positive number indicates the stock of each kind of moon cake （ In 10000 tons ）; The last line gives  N  A positive number indicates the total selling price of each kind of moon cake （ In 100 million yuan ）. Numbers are separated by spaces .

Output format ：

For each group of test cases , Output maximum revenue in one line , In hundred million yuan and accurate to the decimal point 2 position .

sample input ：

3 20
18 15 10
75 72 45

sample output ：

94.50

I wrote it in structure , This question , Main comparison unit price , The higher the unit price , The more you sell , The more profits  

  Code ：

#include<stdio.h>
#include<math.h>
struct
{
    float c,j;
} d[1000],temp;
int main()
{
    float n,m;
    scanf("%f%f",&n,&m);
    int i,k;
    float a[1000],b[1000],l;
    l=0;
    for(i=0; i<n; i++)
    {
        scanf("%f",&a[i]);
    }
    for(i=0; i<n; i++)
    {
        scanf("%f",&b[i]);
    }
    for(k=0; k<n; k++)
    {
        d[k].c=b[k]/a[k];
        d[k].j=a[k];
    }
    for(i=0; i<n-1; i++)
    {
        for(k=0; k<n-1-i; k++)
        {
            if(d[k].c<d[k+1].c)
            {
                temp=d[k];
                d[k]=d[k+1];
                d[k+1]=temp;
            }
        }
    }
    for(i=0; i<n; i++)
    {
        if(d[i].j<=m)
        {
            l=l+d[i].c*d[i].j;
        }
        if(d[i].j>m)
        {
            l=l+d[i].c*m;
            break;
        }
        m=m-d[i].j;
        if(m==0)break;
    }
    printf("%.2f",l);
    return 0;
}

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