FatMouse' Trade 贪心问题~

最新推荐文章于 2019-08-11 08:49:19 发布

TangTWT

最新推荐文章于 2019-08-11 08:49:19 发布

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本文链接：https://blog.youkuaiyun.com/TangTWT/article/details/79220864

这道题目描述了FatMouse用猫粮与猫咪交易JavaBean的故事。FatMouse需要找到最优策略来获取最多的JavaBean。每个房间的JavaBean数量和所需猫粮数量不同，交易比例为a%。输入包含多个测试案例，每个案例给出FatMouse的猫粮总量M和房间数N，以及每个房间的JavaBean和猫粮需求。输出应为每个案例中FatMouse能获取的最大JavaBean数，精确到小数点后三位。解题策略是优先选取比值最大的房间进行交易，并在最后检查是否有剩余猫粮。

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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

Sample Output

13.333
31.500

这是一道简单的贪心问题，比值不同，所以要先用比值大的

最后一步的时候要记得判定是否还有剩余，或者多出了

上代码~

#include<stdio.h>
#include<algorithm>
using namespace std;
struct f
{
    double c;
    int nai,cat;

};
bool pai(f x,f y)
{
    if(x.c>y.c)
    return true;
    else
    return false;
}
int main()
{
    int m,zong,i,j=0,n,b;
    f a[1005];
    while(~scanf("%d%d",&zong,&m))
    {
        double r=0;
            if(zong==-1&&m==-1)
        {
            break;
        }
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&a[i].nai,&a[i].cat);
            a[i].c=a[i].nai*1.0/a[i].cat;
        }
        sort(a,a+m,pai); for(i=0;i<m;i++)        {
            if(zong>a[i].cat||zong==a[i].cat)
            {
                r=a[i].nai+r;
                zong=zong-a[i].cat;
            }
            else
            {
                r=zong*1.0*a[i].c+r;
                zong=0;
                if(zong==0)
                {
                    break;
                }

            }
            if(zong==0)
                {
                    break;
                }

        }
        printf("%.3lf\n",r);
    }

    return 0;
}