2022 Nioke Multi-School Training Session 2 J Question Link with Arithmetic Progression

题目链接

Link with Arithmetic Progression

题目大意

给定一个数组,Let's find a straight line that fits it,Minimum mean squared error is required.I believe that everyone has learned related knowledge in mathematics class or machine learning class.

题解

$$Target=min(\sum _{i=1}^n[a_1+(i-1)\times d-a_i{]}^2)$$
This is the value we are asking for,We need to keep it minimal,We can think about it first$a_1$求出来,很明显,随着$a_1$from negative infinity to positive infinity,TargetThe value first falls and then rises,也就是说$a_1$是一个凹函数,Therefore, we can consider using three points$a_1$,and then considering$a_1$确定的情况下,如何确定$d$.
We deform the formula：
$$=[a_1+(i-1)\times d{]}^2-2\times [a_1+(i-1)\times d]\times a_i+{a_i}^2$$
然后,将带有$d$的合并同类项.
$$=(i-1{)}^2\times d^2+2\times (i-1)\times (a_1-a_i)\times d+{a_1}^2+{a_i}^2$$
According to the knowledge of quadratic functions,我们可以很轻松的确定 $d=\frac{-b}{2\ast (i-1{)}^2}$
然后,就可以进行计算了.
话不多说,上代码：

代码

#include<iostream>
#include<cstring>
#include <cstdio>
#include <cctype>
using namespace std;
#define int long long
#define double long double
const int maxn = 100010;
int w[maxn];
int n;

namespace GTI
{
    
    char gc(void)
       {
    
        const int S = 1 << 16;
        static char buf[S], *s = buf, *t = buf;
        if (s == t) t = buf + fread(s = buf, 1, S, stdin);
        if (s == t) return EOF;
        return *s++;
    }
    int gti(void)
       {
    
        int a = 0, b = 1, c = gc();
        for (; !isdigit(c); c = gc()) b ^= (c == '-');
        for (; isdigit(c); c = gc()) a = a * 10 + c - '0';
        return b ? a : -a;
    }
}
using GTI::gti;
double check(double a1)
{
    
	double a = 0, b = 0;
	for(int i = 1; i <= n; i ++){
    
		a += (i - 1) * (i - 1);
		b += 2 * (i - 1) * (a1 - w[i]);
	}
	double d = - b / (2 * a);
	double res = 0;
	for(int i = 1; i <= n; i ++){
    
		res += (a1 + (i - 1) * d - w[i]) * (a1 + (i - 1) * d - w[i]);
	}
	return res;
}
signed main()
{
    
	int t; t = gti();
	while(t --)
	{
    
		n = gti();
		for(int i = 1; i <= n; i ++) w[i] = gti();
		double l = -1e10, r = 1e10;
		while(r - l > 1e-5) {
    
			double len = r - l;
			double mid_l = l + len / 3, mid_r = r - len / 3;
			if(check(mid_l) >= check(mid_r)) l = mid_l;
			else r = mid_r;
		}
		printf("%.10Lf\n", check(r));
	}
	return 0;
}