uva 10014 Simple calculations（公式推导）

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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

题目大意：给出一个数列的首项、 末项，和中间常量，根据公式a_i = (a_i–1 + a_i+1)/2 – c_i ，求a1项。

解题思路：公式可化简成ai  -   ai-1   = ai+1   -    ai     - 2 * ci,   ai+1 - ai = ai+2 - ai+1 - 2 *ci+1 就可以写成ai - ai-1 + 2 *ci = ai+2 - ai+1 -2 *ci+1, 列出所有项叠项项加后，可得n*（a1 - a0) = an+1 - a0  - sum    (sum可求）

#include<stdio.h>
#include<string.h>
#define N 3005
double a0, an, c[N], a1;
int n;int main(){int t;scanf("%d", &t);while (t--){// Init.memset(c, 0, sizeof(c));// Read.scanf("%d", &n);scanf("%lf%lf", &a0, &an);for (int i = 1; i <= n; i++)scanf("%lf", &c[i]);double sum = 0;for (int i = 1; i <= n; i++)sum = sum + (n + 1 - i) * c[i];a1 = (an + n * a0 - 2 * sum) / (n + 1);printf("%.2lf\n", a1);if (t)printf("\n");}return 0;}

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