UVA 10014 Simple calculations

最新推荐文章于 2018-02-08 08:14:12 发布

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本文介绍了一种基于特定数学规则的数列计算方法，通过给定的起始和结束值及一系列修正系数，推导出数列中指定项的具体值。文章详细解析了计算过程，并提供了实现该算法的C语言代码。

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。

解析：
就是推导公式题

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
……

2an = an-1 + an+1 - 2cn，

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

再把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，然后根据这个公式求出a1输出。

#include<stdio.h>
const int N = 3010;

double c[N];
int main() {
	int t;
	int n;
	double first,end;
	while(scanf("%d",&t) != EOF) {
		while(t--) {
			scanf("%d",&n);
			scanf("%lf",&first);
			scanf("%lf",&end);
			for(int i=1;i<=n;i++) {
				scanf("%lf",&c[i]);
			}

			double sum = n*first + end;

			for(int i=0;i<n;i++) {
				sum -= 2.0*(n-i)*c[i+1];
			}

			printf("%0.2lf\n",sum/(n+1));
			if(t != 0)
				printf("\n");
		}
	}
	return 0;
}