杭電1009--FatMouse' Trade

這是一道簡單的貪心題，題意爲Fatmouse想要儘可能多得守護自己喜歡的JavaBean，所以，i房間得用F[i]d的cat food換得守護J[i]的JavaBean，當JavaBean不足以換的所有時，則根據所有cat food佔所需cat food的比例來乘以i房間JavaBean得到實際能夠得到的JavaBean。


思路很簡單，要用一定的cat food守護儘可能多得JavaBean，必須要房間的JavaBean多並且需要的cat food 少。所以這種涉及兩方面因素的貪心，可以通過兩方面的比率來判斷優先取哪一個。此題應當優先取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（搞清字母含義，不然混亂會加錯，我交了兩遍都是錯在字母搞混含義導致出錯）
                B[k].t=(double)xx/yy;
                k++;
            }
        }//考慮特殊的當房間不需要cat food時，不需要cat food

        sort(B,B+k,cmp);//進行優先排序
       //需要考慮幾種情況，當n=0，時，當m=0時m和n都不爲0時。實際上n=0和m不爲0並且n不爲0可以歸爲一類
        if(m==0)
            printf("%.3lf\n",sum);//直接輸出不需要cat food的JavaBean總和
        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的情況歸爲一類
    }
    return 0;
}