UVa 10014 Simple calculations (数学)

本文主要是介绍UVa 10014 Simple calculations (数学)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

10014 - Simple calculations

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=99&page=show_problem&problem=955

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

11
50.50
25.50
10.15

Sample Output

27.85

思路：

完整代码：

/*0.009s*/#include<cstdio>int main(void)
{int t, n;double a0, an, c, sum;scanf("%d", &t);while (t--){sum = 0.0;scanf("%d%lf%lf", &n, &a0, &an);for (int i = 1; i <= n; i++){scanf("%lf", &c);sum += (n + 1 - i) * c;}printf("%.2f\n", (an + n * a0 - 2 * sum) / (n + 1));if (t)putchar('\n');}return 0;
}

这篇关于UVa 10014 Simple calculations (数学)的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---