FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 13 Accepted Submission(s) : 6

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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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500

程序：

#include <iostream> #include <iomanip>  using namespace std;  struct Trade {      int javaBean, food;      double rate;  };    int main()  {          const int max = 1001;          int catFood, N;          int i, j, f, maxIdx;          double r, get;          Trade record[max], tmp;          while (cin >> catFood >> N && (catFood != -1) && (N != -1)) {                  i = get = 0;                  while (i < N) {                          cin >> j >> f;                          if (j == 0)                                  r = 0.0;                          else if (f == 0)                                  r = static_cast<double>(max);                          else                                  r = 1.0 * j/f;                          record[i].javaBean = j;                          record[i].food = f;                          record[i].rate = r;                          i++;                  }                    for (j = 0; j < N - 1; j++) {                            maxIdx = N - 1;                          for (f = j; f < N; f++){                                  if (record[maxIdx].rate < record[f].rate)                                  maxIdx = f;                          }                           tmp = record[j];                          record[j] = record[maxIdx];                          record[maxIdx] = tmp;  }                 for (i = 0; i < N; i++) {                          f = record[i].food;                          j = record[i].javaBean;                          if (catFood >= f){                                  get += j;                                  catFood -= f;                          }else{                                  r = 1.0 * catFood / f;                                  get += j * r;                                  catFood = 0;                          }                  }cout<<setiosflags(ios::fixed)<<setprecision(3)<<get<<endl;                       }            return 0;  }