【UvaOJ】【基础题目】【Maths - Misc】 10014 - Simple calculations

Simple calculations 

The Problem

There is a sequence of n+2 elements a0, a1,…, an+1 (n <= 3000; -1000 <=  ai 1000). It is known that ai = (ai–1 + ai+1)/2 – ci   for each i=1, 2, ..., n. You are given a0, an+1, c1, ... , cn. 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 an+1 each having two digits after decimal point, and the next n lines contain numbers ci (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 an+1.

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

Sample Input

1

1

50.50

25.50

10.15

Sample Output

27.85

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解题思路：

这题推导过程略麻烦，实际上要经过两次叠加化简

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代码：

#include<iostream>
#include<fstream>
#include<vector>
#include<string>
#include<iomanip>
#include<cmath>
using namespace std;

int main()
{
	ifstream cin("in.txt");
	int N;
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		int n;
		cin >> n;
		cin.get();
		double a0, an1;
		cin >> a0 >> an1;
		
		vector<double> c;
		double tmpc;
		for(int j=0;j<n;j++)
		{
			cin >> tmpc;
			c.push_back(tmpc);
		}
		
		double tmpA = 0;
		for (int j = 1; j <= n; j++)
		{
			tmpA += c[j - 1] * (n - j + 1);
		}
		tmpA *= 2;
		tmpA -= (an1 + n*a0);
		tmpA = tmpA /  (-(1 + n));
		if (abs(tmpA) == 0)
			tmpA = 0;
		cout <<fixed<< setprecision(2) <<tmpA << endl;
	

		if (i != N - 1)
			cout << endl;
	}
}