URAL-1828 approximation by a progression (least squares)

Approximation by a progression

Time Limit: 500MS

Memory Limit: 65536KB

64bit IO Format: %i64d &%i64u

Submit Status

Description

Your is given a sequence of integers  a1,...,  an. Find an arithmetic progression  b1,...,  Bn for which the Value∑ (  AI ?  Bi) 2 is minimal. The elements of the progression can be non-integral.

Input

The first line contains the number  N of elements in the sequence (2≤  N ≤10 4). The second line given the integers  a1,...,  an; Their absolute values do not exceed 10 4.

Output

Output numbers separated with a space:the first term of the required arithmetic progression and its difference, with An absolute or relative error of in most 10 ? 6. It is guaranteed, the answer is unique for all input data.

sample Input

input	output
40 6	0.400 4.900
4-2 -2-2-2	-2 0

Arithmetic progression arithmetic progression of the general formula an=a1+ (n-1) *d, that is an= (a1-d) +n*d is the form of linear equations, and (I,mi) is distributed on both sides of the line, requiring a straight line of k=d,b=a1-d, so think of the least squares, For Y=kx+b, there are k= ((XY) flat--x flat *y Flat)/((x^2) flat-(X flat) ^2), B=y flat--kx flat. It's OK to do it by the formula.

#include <iostream> #include <iomanip>using namespace std;const int maxn = 10005;double A[maxn];int Main () { int N;while (CIN >> N) {for (int i = 1; I <= n; i++) cin >> A[i];d ouble xy=0, x=0, y=0, x2=0;for (int i = 1; I <= N; i++) {xy = xy + a[i] * i;x = x + i;y = y + a[i];x2 = x2 + i*i;}  Double k = (xy/n-(x/n) * (y/n))/(x2/n  -(x/n) * (x/n));d ouble B = y/n-k* (x/n);d ouble a1 = b + k;double d = k;cout <<fixed<<setprecision (6) << A1 << "" <<fixed<<setprecision (6) <<k << Endl;}}

URAL-1828 approximation by a progression (least squares)