UVA10014 Simple calculations【数列】

There is a sequence of n + 2 elements a0, a1, . . . , an+1 (n ≤ 3000; −1000 ≤ ai ≤ 1000). It is known that

for each i = 1, 2, . . . , n. You are given a0, an+1, c1, . . . , cn. Write a program which calculates a1.

Input

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

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

Each test case will be separated by a single line.

Output

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

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

Sample Input

50.50

25.50

10.15

Sample Output

27.85

问题链接：UVA10014 Simple calculations

问题简述：（略）

问题分析：

数列计算问题。

先进行数列通项公式的推导，然后再套公式计算，事情就简单了。

程序说明：（略）

题记：数学问题通常是先进行数学推导，再按照数学公式做计算。

参考链接：（略）

AC的C语言程序如下：

/* UVA10014 Simple calculations */#include <bits/stdc++.h>using namespace std;int main()
{int t;scanf("%d", &t);while (t--) {int n;double a, b, sum = 0;scanf("%d", &n);scanf("%lf%lf", &a, &b);for(int i=0; i<n; i++) {double x;scanf("%lf", &x);sum += (n - i) * x;}printf("%.2f\n", (n * a + b - 2 * sum) / (n + 1.0));if(t)putchar('\n');}return 0;
}

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