HDU-2086

来源：互联网发布：java退出登录代码编辑：程序博客网时间：2024/06/05 17:44

自己推死活没有推出来公式，虽然想到了可以消项但是方法不得当

    /*          因为：Ai=(Ai-1+Ai+1)/2 - Ci,            A1=(A0  +A2  )/2 - C1;           A2=(A1  +  A3)/2 - C2 , ...     =>    A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2)           2[(A1+A2)+(C1+C2)] = A0+A2+A1+A3;           A1+A2 = A0+A3 - 2(C1+C2);     =>    A1+A2 =  A0+A3 - 2(C1+C2)      同理可得：           A1+A1 =  A0+A2 - 2(C1)            A1+A2 =  A0+A3 - 2(C1+C2)           A1+A3 =  A0+A4 - 2(C1+C2+C3)           A1+A4 =  A0+A5 - 2(C1+C2+C3+C4)           ...           A1+An = A0+An+1 - 2(C1+C2+...+Cn)     ----------------------------------------------------- 左右求和          (n+1)A1+(A2+A3+...+An) = nA0 +(A2+A3+...+An) + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)           =>   (n+1)A1 = nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)           =>   A1 = [nA0 + An+1 - 2(nC1+(n-1)C2+...+2Cn-1+Cn)]/(n+1)     */      #include<iostream>      #include<cstdio>      using namespace std;            const int maxn = 3003;      double a[maxn], c[maxn];      int main()      {          int n, i, j;          while (scanf("%d",&n) == 1)          {              scanf("%lf%lf",&a[0],&a[n+1]);              for (i = 1; i <= n; i++)              {                  scanf("%lf",&c[i]);              }              a[1] = n*a[0] + a[n+1];              double sum = 0;              for (i = n, j = 1; i >=1&&j <= n; j++,i--)              {                  sum += i*c[j];              }              a[1] = (a[1] - 2*sum)/(n+1);              printf("%.2lf\n",a[1]);                        }          return 0;      }

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