1047. Simple Calculations_1047. simple calculationstime limit: 1.0 second-优快云博客

本文针对一个特定的数学问题，给出了一种求解方法，并通过一个示例程序详细展示了如何使用C++来解决该问题。文章首先定义了一个含有N+2个元素的数列，然后通过给定的数列首尾值及中间项的差值，计算出了数列的第二个元素。

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1047. Simple Calculations

Time Limit: 1.0 second
Memory Limit: 16 MB

There is a sequence of N + 2 elements a₀, a₁, …, a_N+1 (1 ≤ N ≤ 3000, −2000 ≤ a_i ≤ 2000). 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.

Input

The first line 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.

Output

Output a₁ in the same format as a₀ and a_N+1.

Sample

input	output
1 50.50 25.50 10.15	27.85

根据原式可以得

a[n+1]-a[n]=a[n]-a[n-1]+2*c[n]①

设S[n]=c[1]+c[2]+…+c[n]

对①式叠加相消可以得到a[n+1]-a[1]=a[n]-a[0]+2*S[n]

整理得a[n+1]-a[n]=a[1]-a[0]+2*S[n]②

对②式叠加相消可得到a[n+1]-a[1]=n*(a[1]-a[0])+2*(S[1]+S[2]+…+S[n])

#include <iostream>
using namespace std;

int main()
{
	int i, n;
	double a_0, a_n1, a_1, c[3000], s = 0, temp = 0;
	cin>>n>>a_0>>a_n1;
	for (i=0; i<n; i++)
		cin>>c[i];
	for (i=0; i<n; i++)
	{
		temp += c[i];
		s += temp;
	}
	a_1 = (a_n1 + n * a_0 - 2 * s) / (n + 1);
	printf("%.2f\n", a_1);
}