A1 = ?

A1 = ?

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5116 Accepted Submission(s): 3265

Problem Description

有如下方程：A_i = (A_i-1 + A_i+1)/2 - C_i (i = 1, 2, 3, .... n).
若给出A₀, A_n+1, 和 C₁, C₂, .....C_n.
请编程计算A₁ = ?

Input

输入包括多个测试实例。
对于每个实例，首先是一个正整数n,(n <= 3000); 然后是2个数a₀, a_n+1.接下来的n行每行有一个数c_i(i = 1, ....n);输入以文件结束符结束。

Output

对于每个测试实例，用一行输出所求得的a1(保留2位小数).

Sample Input

1 50.00 25.00 10.00 2 50.00 25.00 10.00 20.00

Sample Output

27.50 15.00

解题思路：

纯粹数学题，找规律:

An = (1/2)An-1 + (1/2)An+1 - Cn

An-1 = (2/3)An-2 + (1/3)An+1 - (2/3)Cn - (4/3)Cn-1

An-2 = (3/4)An-3 + (1/4)An+1 - (1/2)Cn - Cn-1 - (3/2)Cn-2

An-3 = (4/5)An-4 + (1/5)An+1 - (2/5)Cn - (4/5)Cn-1 - (6/5)Cn-2 - (8/5)Cn-3

......

（是不是有点感觉了呢）

接着:

A1 = (n/(n+1))A0 + (1/(n+1))An+1 - (2/(n+1))Cn - (4/(n+1))Cn-1 - ... -(2n/(n+1))C1

     = [ nA0 + An+1 - 2(Cn + 2Cn-1 + 3Cn-2 + ... + nC1) ]/(n+1)

源代码：

#include <stdio.h>
#include <stdlib.h> 
#define MaxSize 3001

int main()
{ 
  int i,n;
  double a1,a0,end;
  double c[MaxSize]; 
  while (scanf("%d",&n)!=EOF)
  { 
    scanf("%lf%lf",&a0,&end);
    for(i=0;i<n;i++)
      scanf("%lf",&c[i]);
    a1 = n* a0 + end ;
    for(i=0;i<n;i++)
      a1-=2*(i+1)*c[n-1-i]; 
    a1 =a1/(1 + n);
    printf ("%.2lf\n",a1);
  } 
  system("pause");
  return 0 ;
}