A1 = ?

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14526    Accepted Submission(s): 8820

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

由题意：

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)

=> 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)

代码：

#include<stdio.h>
int main()
{
	int i,n;
	double c[3005]={0};
	while(scanf("%d",&n)!=EOF)
	{
		double a0,a;
		scanf("%lf%lf",&a0,&a);
		for(i=1;i<=n;i++)
		        scanf("%lf",&c[i]);
		double sum=0,ans,temp=n;
		for(i=1;i<=n;i++)
			sum+=2*(n+1-i)*c[i];
		ans=temp*a0+a-sum;
		ans/=(temp+1);
		printf("%.2lf\n",ans);
	}
	return 0;
}