A1 = ? （数学归纳法）

A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4294    Accepted Submission(s): 2751

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
   
   
   
   

 

已知A1=(A0+A2)/2 - C1, A2=(A1+A3)/2 - C2 , ... =>A1+A2 = (A0+A2+A1+A3)/2 - (C1+C2) =>A1+A2 = A0+A3 - 2(C1+C2) 类似的有： 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) A1+A1 = A0+A2 - 2(C1) (本来就是) ----------------------------------------------------- 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<string.h>
#include<iomanip>
using namespace std;
#define N 3002
double a[N],c[N],sum[N];
int main()
{
	int n;
	while(cin>>n)
	{
		int i;
		double sum1=0;
		//memset(a,0,sizeof(a));
		cin>>a[0]>>a[n];
		memset(sum,0,sizeof(sum));
		//memset(c,0,sizeof(c));
		for(i=1;i<=n;++i)

		{
			cin>>c[i];
			c[i]=2*c[i];
			sum[i]=sum[i-1]+c[i];
			sum1+=sum[i];
		}
		cout<<setiosflags(ios::fixed)<<setprecision(2)<<(n*a[0]+a[n]-sum1)/(n+1)<<endl;
	}
return 0;
}