A1 = ?

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A1 = ?

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8204 Accepted Submission(s): 5071

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A ₀, A _n+1, 和 C ₁, C ₂, .....C _n.
请编程计算A ₁ = ?

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a ₀, 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

Sample Output

  
   27.50
15.00

Source

2006/1/15 ACM程序设计期末考试

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//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;
int main ()
{
    int i,j,n;
    double sum;
    while (~scanf("%d",&n))
    {
        sum=0;
        double a[3500],c[3500],d;
        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];
        for (i = n, j = 1; i >=1&&j <= n; j++,i--)
            sum += i*c[j];
        a[1] = (a[1] - 2*sum)/(n+1);
        //sum+=c[i];
        //cout<<a[1]<<endl;
        printf ("%.2lf\n",a[1]);
    }
    return 0;
}