HDU ACM 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26437    Accepted Submission(s): 8487

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

 

Recommend

JGShining

 

#include<stdio.h>
#include<string.h>
int J[1002], F[1002];
double JF[1002];
int main()
{
    int j, m, n, i;
    double sum, temp;
    while(scanf("%d%d", &m, &n) != EOF && m != -1 && n != -1)
    {
        memset(J, 0, sizeof(J));
        memset(F, 0, sizeof(F));
        memset(JF,0, sizeof(JF));
        for(i=1; i<=n; ++i)  //从1开始 
        {
            scanf("%d%d", &J[i], &F[i]);
            JF[i] = J[i]*1.0/F[i];
        }

        for(i=2; i<=n; ++i)
        {
            if(JF[i] >= JF[i-1])
            {
                JF[0] = JF[i];
                J[0] = J[i];
                F[0] = F[i];
                
                JF[i] = JF[i-1];
                J[i] = J[i-1];
                F[i] = F[i-1];                
                
                for(j=i-2; j>=0; --j)
                {
                    if(JF[0] >= JF[j])
                    {
                        JF[j+1] = JF[j];
                        J[j+1] = J[j];
                        F[j+1] = F[j];                        
                    }
                    else {
                            JF[j+1] = JF[0]; 
                            J[j+1] = J[0]; 
                            F[j+1] = F[0]; 
                            break;
                        }
                }
            }
         } 
                  
         sum = 0;
         temp = m;
         for(i=1; i<=n; ++i)
         {
             temp = temp - F[i];
             if(temp >= 0) sum = sum + 1.0*J[i];
             else {sum = sum + (temp + F[i])*JF[i]; break;} 
            
         }
         printf("%.3f\n", sum);
        
        
    }   
    return 0; 
}

结题报告：

1AC的感觉确实比较好，但是到底是因为什么才ＡＣ的，我却不知道怎么回答，天知道提交之后会有什么后果，我只知道有人告诉我这题耐心做还是很容易做出来的，所以就顺理成章的耐心地做出来了，还好，让我知道了耐心、坚持、自信的重要性。

肥老鼠换东西，比率ａ％就要看你给那只猫多好了，对于每一个房间，给你的提成或者说利润＝J[i]/F[i],　所以刚开始就要计算各房间的利润然后排序，然后问你了：你会将钱投资给利润高的还是利润低的？