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  • PAT Basic Level 1020

    AC代码

     1 #include <stdio.h>
     2 #include <stdlib.h>
     3 struct mk
     4 {
     5     double storage;
     6     double sumPrice;
     7     double perPrice;
     8 };
     9 int cmp(const void *a,const void *b)           //用于qsort按照单价排序 
    10 {
    11     struct mk m1 = *(struct mk*)a;
    12     struct mk m2 = *(struct mk*)b;
    13     return m2.perPrice > m1.perPrice?1:-1;          //仔细思考这一行 
    14 }
    15 int main ()
    16 {
    17     int num;
    18     double need;
    19     double ret = 0.0;
    20     scanf("%d %lf",&num,&need);
    21     struct mk list[1100];
    22     int i;
    23     for(i = 0;i < num;i++)
    24     {
    25         scanf("%lf",&list[i].storage);
    26     }
    27     for(i = 0;i < num;i++)
    28     {
    29         scanf("%lf",&list[i].sumPrice);
    30     }
    31     for(i = 0;i < num;i++)                       //以上为输入 
    32     {
    33         list[i].perPrice = list[i].sumPrice/list[i].storage;    
    34     }
    35     qsort(list,num,sizeof(list[0]),cmp);
    36     i = 0;
    37     while(need != 0)
    38     {
    39         if(need > list[i].storage)          //需求大于库存 
    40         {
    41             need = need - list[i].storage;
    42             ret = ret + list[i].sumPrice; 
    43             if( i == num-1)                     //全部用完还是没满足需求 
    44             {
    45                 need = 0;
    46             }
    47         }
    48         else                       //需求小于库存 
    49         {
    50             ret = ret + list[i].perPrice * need;
    51             need = 0;
    52         }
    53         i++;
    54     }
    55     printf("%.2lf
    ",ret);                   
    56     return 0 ;
    57 }
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  • 原文地址:https://www.cnblogs.com/Ponytai1/p/5976507.html
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