2019暑假牛客多校赛C.Euclidean Distance(暴力+数学)

最新推荐文章于 2024-07-16 21:55:30 发布

原创最新推荐文章于 2024-07-16 21:55:30 发布 · 224 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#暴力+数学

数论专栏收录该内容

2 篇文章

订阅专栏

本文探讨了在n维实空间中，给定点A坐标为(a1/m,...,an/m)，寻找另一点P=(p1,...,pn)，使得点P与A之间的欧氏距离平方最小。在约束条件下，p1+...+pn=1且pi≥0，文章提供了一个通过拉格朗日乘子法解决此问题的算法实现。

C.Euclidean Distance

传送门

题目描述
Bobo has a point A in the n dimension real space $\mathbb{R}^n$ , whose coodinate is $(a_1 / m, a_2 / m, \dots, a_n / m)$ and m are both integers. He wants to find another point P = $(p_1, p_2, \dots, p_n)$ meeting the following requirements. $p_1, p_2, \dots, p_n \in \mathbb{R}$ .That is, they are real numbers.

$p_1, p_2, \dots, p_n \geq 0$
$p_1 + p_2 + \dots + p_n = 1$
The (squared) Euclidean distance between P and A, which is $\|A - P\|_2^2 = \sum_{i = 1}^n (a_i / m - p_i)^2$ , is minimized.

It can be proved the minimum is always a rational number. Print the squared distance in fraction. Note to print an integer n as n instead of n/1.
输入描述:
The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains two integers n and m.
The second line contains n integers $a_1, a_2, \dots, a_n$ $\leq n \leq 10^4$
$\leq m \leq 10^3$ $\leq a_i \leq m$
The sum of n does not exceed $\times 10^5$
.
输出描述:
For each test case, print a fraction which denotes the result.
示例1
输入
1 1
0
2 3
1 2
3 10
1 -2 3
输出
1
0
16/75

题目大意：
给定 m 和 n 个整数 ai， $m≤a_i≤m$ ，求 $\|A - P\|_2^2 = \sum_{i = 1}^n (a_i / m - p_i)^2$ 在约束条件 $p_1 + p_2 + \dots + p_n = 1$ , $p_1, p_2, \dots, p_n \geq 0$ 下的最小值。

思路：
看大佬博客
听说标程是拉格朗日乘子法做的，但是查了几篇博客都没有用拉格朗日乘子法。

ac_code:

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 7;

int n, m;
LL a[maxn];

int main() {
    while(~scanf("%d%d",&n,&m)){
        for(int i = 1;i<=n;++i){
            scanf("%lld",&a[i]);
        }
        sort(a+1,a+n+1,[](int a,int b){return a>b;});
        LL idx=n;
        LL las=m;
        for(int i=1;i<n;++i){
            if(i*abs(a[i+1]-a[i])>las){
                idx=i;
                break;
            }else{
                las-=i*abs(a[i+1]-a[i]);
            }
        }
        LL num1=1LL*(idx*a[idx]-las)*(idx*a[idx]-las);
        LL num2=1LL*idx*m*m;
        for(int i=idx+1;i<=n;++i){
            num1+=1LL*a[i]*a[i]*idx;
        }
        LL tmp=__gcd(num1,num2);
        num1/=tmp,num2/=tmp;
        if(num2==1) printf("%lld\n",num1);
        else printf("%lld/%lld\n",num1,num2);
    }
    return 0;
}