UVA 10014 Simple calculations

 Simple calculations 

The Problem

There is a sequence of n+2 elements a₀, a₁,…, a_n+1 (n <= 3000; -1000 <=  a_i 1000). It is known that a_i = (a_i–1 + a_i+1)/2 – c_i   for each i=1, 2, ..., n. You are given a₀, a_n+1, c₁, ... , c_n. Write a program which calculates a₁.

The Input

The first line is the number of test cases, followed by a blank line.

For each test case, the first line of an input file contains an integer n. The next two lines consist of numbers a₀ and a_n+1 each having two digits after decimal point, and the next n lines contain numbers c_i (also with two digits after decimal point), one number per line.

Each test case will be separated by a single line.

The Output

For each test case, the output file should contain a₁ in the same format as a₀ and a_n+1.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

1150.5025.5010.15

Sample Output

27.85

题意：。。根据题目给的公式。输入了a0, a(n + 1)..和n个Ci。要根据公式推出a1.。。

就是推导公式。。。

将i等于1、 2 、3 ...n带入原式子。。。

得到

 2a₁ = a₀ + a₂ - 2c₁

   2a₂ = a₁ + a₃ - 2c2

   2a₃ = a₂ + a₄ - 2c₃

   …… ……   ……

   2a_n = a_n-1 + a_n+1 - 2c_n，

可以得到 a1 - a0 + 2(c1 +c2 +... +cn) = a(n + 1) - a(n).

在把n = 1， 2， 3 ，4 ，5 ，6...n 。。带入

得到a1 - a0 + 2(c1) = a2 - a1;

        a1 - a0 + 2(c1 + c2) = a3 - a2;

。。。。

a1 - a0 + 2(c1 + c2 + c3 +...cn) = a(n+1) - a(n);

然后把上式子全部叠加。。可以得到n * (a1 - a0) + 2(n*c1 + (n -1) * c2 + (n -2) * c3+ .... 2*cn - 1 + cn) = a(n + 1) - a1;

这样除了a1 其他全是已知量了。 - - 好麻烦啊。。。然后根据这个公式求出a1输出。

 

#include <stdio.h>
#include <string.h>

int t;
int n;
double star, end;
double c[3005];
double a1;
int main()
{
    scanf("%d", &t);
    while (t --)
    {
	memset(c, 0 ,sizeof(c));
	scanf("%d", &n);
	    scanf("%lf%lf",&star, &end);
	for (int i = 1; i <= n; i ++)
	    scanf("%lf", &c[i]);
	a1 = (end + n * star);
	for (int i = 1; i <= n; i ++)
	{
	    a1 -= 2 * c[i] * (n + 1 - i);
	}
	a1 /= n + 1;
	printf("%.2lf
", a1);
	if (t)
	    printf("
");
    }
    return 0;
}