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

 

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JGShining

 

#include
      
       #include
       
        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？