HDU 2086 A1 = ?（规律）

A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5932    Accepted Submission(s): 3721

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

   
   
   
   

    
    
    
    1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    27.50
15.00
   
   
   
   

 

Source

2006/1/15 ACM程序设计期末考试

 

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n=1;2A1=A0+A2-2C1

n=2;3A1=2A0+A3-4C1-2C2

n=3;4A1=3A0+A4-6C1-4C2-2C3

n=5;5A1=4A0+A5-8C1-6C2-4C3-2C4;

 

#include<stdio.h>
int main(){
    double a0,a1,ai,c[3010];
    int n;
    while(~scanf("%d",&n)){
        scanf("%lf%lf",&a0,&ai);
        a1=n*a0+ai;
        int k=n;
        for(int i=1;i<=n;i++){
            scanf("%lf",&c[i]);
            a1=a1-k*2*c[i];
            k--;
        }
        printf("%.2lf\n",a1/(n+1));
    }
    return 0;
}