URAL-1828 Approximation by a Progression (Least Square Method)

URAL-1828 Approximation by a Progression (Least Square Method)
Approximation by a Progression

Time Limit:500 MS

Memory Limit:65536KB

64bit IO Format:% I64d & % I64u

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Description

Your are given a sequence of integers A1 ,..., An. Find an arithmetic progression B1 ,..., BnFor which the value Σ ( Ai? Bi) 2 is minimal. The elements of the progression can be non-integral.

Input

The first line contains the number NOf elements in the sequence (2 ≤ N≤ 10 4). In the second line you are given the integers A1 ,..., An; Their absolute values do not exceed 10 4.

Output

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

Sample Input

Input	Output
40 6 10 15	0.400 4.900
4-2 -2 -2 -2	-2 0

The general an = a1 + (n-1) * d of the arithmetic ssion proportional difference sequence, that is, an = (a1-d) + n * d is the form of a linear equation, while (I, mi) is distributed on both sides of the straight line, requiring a straight line k = d, B = a1-d, so think of the least squares, on Y = kX + B, k = (XY) flat -- X flat * Y flat)/(X ^ 2) flat -- (X flat) ^ 2), B = Y flat -- kX flat. Just find the formula.

#include
 
  #include
  
   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];double 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));double b = y / n - k*(x / n);double a1 = b + k;double d = k;cout <