hdu 1009 FatMouse' Trade

解决HDU 1009问题：猫鼠交易最大Java豆获取量

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本博客详细介绍了如何通过算法解决HDU 1009问题，即在猫鼠交易场景中，通过优化算法计算出最多可以获取的Java豆数量。涉及输入数据解析、排序算法应用、贪心策略实现等核心步骤，旨在提高读者对算法优化的理解和实践能力。

链接:hdu 1009

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

Sample Output

  
   13.333
31.500

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
struct stu
{
    int j,f;
    double a;
};
int cmp(struct stu a,struct stu b)
{
    return a.a>b.a;
}
int main()
{
    struct stu s[1010];
    int i,n,m;
    double sum;
    while(scanf("%d%d",&m,&n)!=EOF){
        if(m==-1&&n==-1)
            break;
        for(i=1;i<=n;i++){
            scanf("%d%d",&s[i].j,&s[i].f);
            s[i].a=(double)s[i].j/s[i].f;
        }
        sort(s+1,s+n+1,cmp);
        sum=0;
        for(i=1;i<=n;i++){
            if(m>=s[i].f){
                sum+=s[i].j;
                m-=s[i].f;
            }
            else{
                sum+=s[i].a*m;
                m=0;
            }
            if(m==0)
                break;
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}