• POJ 1701


      1 /*
      2 Every tenant went up N floors would make the dissatisfied degree rise N * a + 0.5 * N * (N - 1) degrees,
      3 every tenant went down N floors would make the dissatisfied degree rise N * b + 0.5 * N * (N - 1) degrees. 
      4 
      5   △= Dp+1 – Dp
      6 = ∑1=<i<=p Ki*(b+p-i)) - ∑p<i<=n Ki*(a+i-p-1)
      7 =  ∑1=<i<=n Ki * i +(b+p)*∑1=<i<=p Ki   -     (a-p-1)∑p<i<=n Ki
      8 */
      9 
     10 #include <iostream>
     11 #define MAXN  10001
     12 using namespace std;
     13 
     14 __int64 _m[MAXN+1];
     15 int k[MAXN+1];
     16 __int64 p_2[MAXN+1];
     17 __int64 p_3[MAXN+1];
     18 int k_1[MAXN];
     19 int k_2[MAXN];
     20 
     21 
     22 int main()
     23 {
     24     //freopen("acm.acm","r",stdin);
     25     int test;
     26     int n;
     27     int i;
     28     int j;
     29     int  p_1;
     30     int a;
     31     int b;
     32     scanf("%d",&test);
     33     while(test --)
     34     {
     35         scanf("%d",&n);
     36         scanf("%d",&a);
     37         scanf("%d",&b);
     38     
     39         for(i = 1; i <= n; ++ i)
     40         {
     41             scanf("%d",&k[i]);
     42         }
     43 
     44         for(i = 1; i <= n; ++ i)
     45         {
     46             k_1[i] = k_2[i] = k[i];
     47         }
     48 
     49         for(i = n-1; i >=1; -- i)
     50         {
     51             k_2[i] += k_2[i+1];
     52         }
     53 
     54         for(i = 2; i <= n; ++i)
     55         {
     56             k_1[i] += k_1[i-1];
     57         }
     58 
     59         p_1 = 0;
     60         for(i = 1; i <= n; ++ i)
     61         {
     62             p_1 += k[i]*i;
     63         }
     64 
     65         for(i = 1; i <= n; ++ i)
     66         {
     67             p_2[i] = (b + i)*k_1[i];
     68         }
     69 
     70         for(i = 1; i <= n; ++ i)
     71         {
     72             k_2[i] -= k[i];    
     73         }
     74                 
     75 
     76         copy(k_2,k_2+n+1,p_3);
     77     
     78         for(i = 1; i <= n; ++ i)
     79         {
     80             p_3[i] *= (a-i-1);
     81         }
     82         _m[1] = 0;
     83         for(i = 2; i <= n; ++ i)
     84         {
     85             _m[1] += (i-1)*a+((i-1)*(i-2)/2);
     86         }
     87 
     88         for(i = 2; i <= n; ++ i)
     89         {
     90             _m[i] = _m[i-1] - p_1 + p_2[i-1] - p_3[i-1];
     91         }
     92 
     93         __int64 min = 999999999999999;
     94 
     95         int place;
     96     
     97         for(i = 1; i <= n; ++ i)
     98         {
     99             if(_m[i] < min)
    100             {
    101                 min = _m[i];
    102                 place = i;
    103             }
    104         }
    105         cout<<place<<endl;
    106     }
    107 }

    关注我的公众号,当然,如果你对Java, Scala, Python等技术经验,以及编程日记,感兴趣的话。 

    技术网站地址: vmfor.com

  • 相关阅读:
    Mootools遮罩层练习(原为网上的jquery写法)
    Notepad++ c编译环境 64
    Apache添加多端口
    数据库权限分配(远程共享数据库)(mysql)
    Zend studio13 导入已有php文件夹
    按钮js跳转到非表单提交页
    Sql case when 小例
    php 获取数组中的key值
    knit使用安装 ecstore knit位置 及调用路径
    python数据库操作
  • 原文地址:https://www.cnblogs.com/gavinsp/p/4566506.html
Copyright © 2020-2023  润新知