商人的诀窍

Problem Description

E_star和von是中国赫赫有名的两位商人，俗话说的好无商不奸，最近E_star需要进一批苹果。可是他需要的苹果只有von才有，von的苹果都存在他的传说中很牛叉的仓库里，每个仓库都存了不同种类的苹果，而且每个仓库里的苹果的价钱不同。如果E_star想要买仓库i里的所有重量为f[i]的苹果他必须付m[i]的金钱。E_star开着他的传说中的毛驴车去拉苹果，而且他只带了N些金钱。E_star作为传说中的奸商希望用它所带的N金钱得到重量最多的苹果。你作为他最好的朋友，所以他向你求出帮助。希望你能帮忙计算出他能买到最多的苹果（这里指重量最大）。并输出最大重量。

提示：这里仅考虑仓库里苹果的重量，不考虑个数。

Input

第一行包括两个非负整数N,M（分别代表E_star带的金币数,von盛苹果的仓库数量，不超过50）。

接下来有有M行，每行包括两个数非负整数f[i]和m[i]分别表示第i仓库里存有重量为f[i]的苹果，如果将所有苹果买下要花费m[i]的金钱，E_star不必非要将每个仓库的苹果全部买下。

当M,N二者中任一为-1时结束。

Output

E_star用N的金币所能买到的最大重量的苹果的重量。结果保留三位小数。

Sample Input

Sample Output

13.333
31.500

Hint

Source

E_st

错误的：

 
     01 
     #include <stdio.h>

 
     02 
     struct node

 
     03 
     {

 
     04 
         int E , von;

 
     05 
         double jia;

 
     06 
     }shop[10001];

 
     07 
     void pai(struct node shop [] , int l ,int r)

 
     08 
     {

 
     09 
         int i = l , j = r;

 
     10 
         struct node k = shop[l];

 
     11 
         if (l >= r)

 
     12 
         {

 
     13 
             return ;

 
     14 
         }

 
     15 
         while(i < j)

 
     16 
         {

 
     17 
             while (i < j && shop[j].jia <= k.jia)

 
     18 
             {

 
     19 
                 j--;

 
     20 
             }

 
     21 
             shop[i] = shop[j];

 
     22 
             while(i < j && shop[i].jia >= k.jia)

 
     23 
             {

 
     24 
                 i++;

 
     25 
             }

 
     26 
             shop[j] = shop[i];

 
     27 
         }

 
     28 
         shop[i] = k;

 
     29 
         pai(shop , i + 1 , r);

 
     30 
         pai(shop , l , i - 1);

 
     31 
     }

 
     32 
     int main ()

 
     33 
     {

 
     34 
         int n , m , i;

 
     35 
         while(~scanf("%d%d" , &n ,&m) &&(n != -1 || m!= -1))

 
     36 
         {

 
     37 
             for (i = 0 ; i < m ; i++)

 
     38 
             {

 
     39 
                 scanf("%d%d" , &shop[i].E , &shop[i].von);

 
     40 
                 shop[i].jia = shop[i].E *(1.000 / shop[i].von);

 
     41 
             }

 
     42 
             pai(shop , 0 , m - 1);

 
     43 
             i= 0;

 
     44 
             double sum = shop[i].E;

 
     45 
             int z = shop[0].von;

 
     46 
             i = 1;

 
     47 
             while(z  < n)

 
     48 
             {

 
     49 
                 if (z + shop[i].von <= n)

 
     50 
                 {

 
     51 
                     sum += shop[i].E;

 
     52 
                     z += shop[i].von;

 
     53 
                 }

 
     54 
                 else

 
     55 
                 {

 
     56 
                     sum += (n - z) * shop[i].jia ;

 
     57 
                    break;

 
     58 
                 }

 
     59 
                 i++;

 
     60 
             }

 
     61 
             printf("%.3lf\n" , sum);

 
     62 
         }

 
     63 
         return 0;

 
     64 
     }

65

66

正确的：

view source
print?

 
      01 
      #include <stdio.h> 
     
      02 
      struct node 
     
      03 
      { 
     
      04 
          int E , von; 
     
      05 
          double jia; 
     
      06 
      }shop[10001]; 
     
      07 
      void pai(struct node shop [] , int l ,int r) 
     
      08 
      { 
     
      09 
          int i = l , j = r; 
     
      10 
          struct node k = shop[l]; 
     
      11 
          if (l >= r) 
     
      12 
          { 
     
      13 
              return ; 
     
      14 
          } 
     
      15 
          while(i < j) 
     
      16 
          { 
     
      17 
              while (i < j && shop[j].jia <= k.jia) 
     
      18 
              { 
     
      19 
                  j--; 
     
      20 
              } 
     
      21 
              shop[i] = shop[j]; 
     
      22 
              while(i < j && shop[i].jia >= k.jia) 
     
      23 
              { 
     
      24 
                  i++; 
     
      25 
              } 
     
      26 
              shop[j] = shop[i]; 
     
      27 
          } 
     
      28 
          shop[i] = k; 
     
      29 
          pai(shop , i + 1 , r); 
     
      30 
          pai(shop , l , i - 1); 
     
      31 
      } 
     
      32 
      int main () 
     
      33 
      { 
     
      34 
          int n , m , i; 
     
      35 
          while(~scanf("%d%d" , &n ,&m) &&(n != -1 || m!= -1)) 
     
      36 
          { 
     
      37 
              for (i = 0 ; i < m ; i++) 
     
      38 
              { 
     
      39 
                  scanf("%d%d" , &shop[i].E , &shop[i].von); 
     
      40 
                  shop[i].jia = shop[i].E *(1.000 / shop[i].von); 
     
      41 
              } 
     
      42 
              pai(shop , 0 , m - 1); 
     
      43 
                
      44 
              double sum = 0; 
     
      45 
             for(i = 0; i < m; i++) 
     
      46 
             { 
     
      47 
                 if(n > shop[i].von) 
     
      48 
                 { 
     
      49 
                     sum += shop[i].E; 
     
      50 
                     n -= shop[i].von; 
     
      51 
                 } 
     
      52 
        
      53 
        
      54 
                 else 
     
      55 
                 { 
     
      56 
                     sum += 1.0 * n* shop[i].jia; 
     
      57 
                     break; 
     
      58 
                 } 
     
      59 
             } 
     
      60 
        
      61 
              printf("%.3lf\n" , sum); 
     
      62 
          } 
     
      63 
          return 0; 
     
      64 
      }

view source
print?

 
       01 
       #include <stdio.h> 
      
       02 
       struct node 
      
       03 
       { 
      
       04 
           int E , von; 
      
       05 
           double jia; 
      
       06 
       }shop[10001]; 
      
       07 
       void pai(struct node shop [] , int l ,int r) 
      
       08 
       { 
      
       09 
           int i = l , j = r; 
      
       10 
           struct node k = shop[l]; 
      
       11 
           if (l >= r) 
      
       12 
           { 
      
       13 
               return ; 
      
       14 
           } 
      
       15 
           while(i < j) 
      
       16 
           { 
      
       17 
               while (i < j && shop[j].jia <= k.jia) 
      
       18 
               { 
      
       19 
                   j--; 
      
       20 
               } 
      
       21 
               shop[i] = shop[j]; 
      
       22 
               while(i < j && shop[i].jia >= k.jia) 
      
       23 
               { 
      
       24 
                   i++; 
      
       25 
               } 
      
       26 
               shop[j] = shop[i]; 
      
       27 
           } 
      
       28 
           shop[i] = k; 
      
       29 
           pai(shop , i + 1 , r); 
      
       30 
           pai(shop , l , i - 1); 
      
       31 
       } 
      
       32 
       int main () 
      
       33 
       { 
      
       34 
           int n , m , i; 
      
       35 
           while(~scanf("%d%d" , &n ,&m) &&(n != -1 || m!= -1)) 
      
       36 
           { 
      
       37 
               for (i = 0 ; i < m ; i++) 
      
       38 
               { 
      
       39 
                   scanf("%d%d" , &shop[i].E , &shop[i].von); 
      
       40 
                   shop[i].jia = shop[i].E *(1.000 / shop[i].von); 
      
       41 
               } 
      
       42 
               pai(shop , 0 , m - 1); 
      
       43 
                 
       44 
               double sum = 0; 
      
       45 
              for(i = 0; i < m; i++) 
      
       46 
              { 
      
       47 
                  if(n > shop[i].von) 
      
       48 
                  { 
      
       49 
                      sum += shop[i].E; 
      
       50 
                      n -= shop[i].von; 
      
       51 
                  } 
      
       52 
         
       53 
         
       54 
                  else 
      
       55 
                  { 
      
       56 
                      sum += 1.0 * n* shop[i].jia; 
      
       57 
                      break; 
      
       58 
                  } 
      
       59 
              } 
      
       60 
         
       61 
               printf("%.3lf\n" , sum); 
      
       62 
           } 
      
       63 
           return 0; 
      
       64 
       }