hdu-1009 FatMouse' Trade 贪心算法

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

 

题意：一只肥老鼠和猫交易的故事。。。

一只肥老鼠有m磅猫粮，准备与守卫仓库的猫交换它最喜欢的食物。第一行输入两个整数m和n，表示老鼠有m磅猫粮，猫的仓库有n个房间，接下来是n组数据，每组数据包括两个整数老鼠和猫交换的粮食数量，如第一组老鼠可以用2磅猫粮交换7磅粮食。计算出老鼠可以得到的最多粮食磅数。

毫无疑问的贪心算法，既然老鼠要得到最多磅数的粮食，则老鼠必须用最少的猫粮换最多的粮食。

现将每组数据排序，按每磅猫粮可以交换的粮食从多到少排序。如第一组测试数据，第一组数据中每磅猫粮可以叫唤3.5磅粮食，第二组中每磅猫粮可以交换1.333磅粮食，第三组中每磅猫粮可以交换2.5磅粮食，排序后的数组为7 2，5 2，4 3，老鼠的五磅猫粮前四磅交换了7+5=12磅粮食，最后一磅交换了最后一组中4的1/3磅粮食，所以老鼠交换的总粮食磅数为13.333磅粮食。

ps.这里可以优化的是，如果老鼠的猫粮总量大于所有房间内猫粮总量之和，说明老鼠可以拿到所有的粮食，直接输出粮食之和即可。

AC代码

#include 
#include 
double a[1010][2];
int main()
{
    double temp;
    int i,m,n,k;
    double sum1,sum;
    while (scanf("%d %d",&m,&n)!=EOF) {
        sum=0;sum1=0;
        memset(a, 0, sizeof(a));
        if(m==-1&&n==-1)
            return 0;
        for(i=0;i=sum){
            printf("%.3lf\n",sum1);
            continue;
        }
        sum=0;sum1=0;
        for(i=0;ia[i][0]/a[i][1])
                {
                    temp=a[k][0];
                    a[k][0]=a[i][0];
                    a[i][0]=temp;
                    temp=a[k][1];
                    a[k][1]=a[i][1];
                    a[i][1]=temp;
                }

        for(i=0;im){
                sum=sum-a[i][1];
                break;
            }
            sum1=sum1+a[i][0];
        }
        printf("%.3lf\n",sum1+1.0*a[i][0]*(m-sum)/a[i][1]);
    }
}