• USACO Training 1.4 Ski Course Design


    Farmer John has N hills on his farm (1 <= N <= 1,000), each with an integer elevation in the range 0 .. 100. In the winter, since there is abundant snow on these hills, FJ routinely operates a ski training camp.

    Unfortunately, FJ has just found out about a new tax that will be assessed next year on farms used as ski training camps. Upon careful reading of the law, however, he discovers that the official definition of a ski camp requires the difference between the highest and lowest hill on his property to be strictly larger than 17. Therefore, if he shortens his tallest hills and adds mass to increase the height of his shorter hills, FJ can avoid paying the tax as long as the new difference between the highest and lowest hill is at most 17.

    If it costs x^2 units of money to change the height of a hill by x units, what is the minimum amount of money FJ will need to pay? FJ can change the height of a hill only once, so the total cost for each hill is the square of the difference between its original and final height. FJ is only willing to change the height of each hill by an integer amount.

    PROGRAM NAME: skidesign

    INPUT FORMAT:

    Line 1:The integer N.
    Lines 2..1+N:Each line contains the elevation of a single hill.

    SAMPLE INPUT (file skidesign.in):

    5
    20
    4
    1
    24
    21
    

    INPUT DETAILS:

    FJ's farm has 5 hills, with elevations 1, 4, 20, 21, and 24.

    OUTPUT FORMAT:

    The minimum amount FJ needs to pay to modify the elevations of his hills so the difference between largest and smallest is at most 17 units.

    SAMPLE OUTPUT (file skidesign.out):

    18
    

    OUTPUT DETAILS:

    FJ keeps the hills of heights 4, 20, and 21 as they are. He adds mass to the hill of height 1, bringing it to height 4 (cost = 3^2 = 9). He shortens the hill of height 24 to height 21, also at a cost of 3^2 = 9.

    1. /*
    2. ID:choiyin1
    3. PROG:skidesign
    4. LANG:C++
    5. */
    6. #include
    7. using namespace std;
    8. int N;
    9. int h[1001];
    10. #define MIN(a,b) ((a)>(b)?(b):(a))
    11. int main(int argc,char *argv[])
    12. {
    13. freopen("skidesign.in","r",stdin);
    14. freopen("skidesign.out","w",stdout);
    15. scanf("%d",&N);
    16. int Max=0;
    17. int tmp;
    18. for(int i=0;i
    19. {
    20. scanf("%d",&tmp);
    21. if(Max
    22. h[i]=tmp;
    23. }
    24. int ans=0x3f3f3f3f;
    25. for(int i=0;i<=Max;++i)
    26. {
    27. int sum=0;
    28. for(int j=0;j
    29. {
    30. if(h[j]17)
    31. {
    32. int c=(i-h[j]-17)*(i-h[j]-17);
    33. sum+=c;
    34. }
    35. if(h[j]>i)
    36. {
    37. int c= h[j]-i;
    38. sum+=c*c;
    39. }
    40. }
    41. ans=MIN(sum,ans);
    42. }
    43. printf("%d\n",ans);
    44. }

  • 相关阅读:
    SpringBoot框架分层(View层、Controller层、Service层、Mapper层、pojo层)
    Mabitys总结
    内存规整how
    sql 判断表数据是否存在另一个表中
    什么是分页?如何使用分页?
    【leetcode】【2022/9/13】670. 最大交换
    linux命令:umask
    opencv形状目标检测
    经纬恒润48V BMS助力Stellantis量产落地
    二维码智慧门牌管理系统升级解决方案:轻松实现辖区范围门址统计
  • 原文地址:https://blog.csdn.net/GeekAlice/article/details/127037928