杭电1009--FatMouse' Trade

这是一道简单的贪心题，题意为Fatmouse想要尽可能多得守护自己喜欢的JavaBean，所以，i房间得用F[i]d的cat food换得守护J[i]的JavaBean，当JavaBean不足以换的所有时，则根据所有cat food占所需cat food的比例来乘以i房间JavaBean得到实际能够得到的JavaBean。


思路很简单，要用一定的cat food守护尽可能多得JavaBean，必须要房间的JavaBean多并且需要的cat food 少。所以这种涉及两方面因素的贪心，可以通过两方面的比率来判断优先取哪一个。此题应当优先取J[i]/F[i]大的。

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41918    Accepted Submission(s): 13938

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

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#include<stdio.h>
#include<algorithm>
using namespace std;
struct cat
{
    int j;//javabean
    int f;//cat fod
    double t;
}B[1005];
typedef struct cat cat;
bool cmp(cat a,cat b)
{
    return a.t>b.t;
}
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        int i;
        double sum=0;
        int k=0;
        int xx,yy;
        if(m==-1&&n==-1)
            break;
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&xx,&yy);
            if(yy==0)
                sum+=xx;
            else
            {
                B[k].j=xx;//javabean
                B[k].f=yy;//cat food（搞清字母含义，不然混乱会加错，我交了两遍都是错在字母搞混含义导致出错）
                B[k].t=(double)xx/yy;
                k++;
            }
        }//考虑特殊的当房间不需要cat food时，不需要cat food

        sort(B,B+k,cmp);//进行优先排序
       //需要考虑几种情况，当n=0，时，当m=0时m和n都不为0时。实际上n=0和m不为0并且n不为0可以归为一类
        if(m==0)
            printf("%.3lf\n",sum);//直接输出不需要cat food的JavaBean总和
        else
        {
            int temp=m;
            for(i=0;i<k;i++)
            {
                if(temp<=B[i].f)
                    sum+=B[i].t*(double)temp;
                else
                    sum+=(double)B[i].j;
                temp-=B[i].f;
                if(temp<=0)
                    break;
            }
            printf("%.3lf\n",sum);
        }//n=0和m!=0&&n!=0的情况归为一类
    }
    return 0;
}