题意：求两个手环任意旋转对应位置的差值+c的平方最小
设b旋转到k最小，那么先将b扩张一倍构成一圈，那么答案式子就是
将这个式子展开一下，事情就变得有趣了起来
这个式子将a[ ]翻转可以化成卷积形式
直接套上一个FFT就可以了
然后枚举C就行了（C的范围比较小，主要是写起来容易，追求效率可以用求根公式算出这个二次函数取得最小值的时候，C的大小）
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-9
#define fi first
#define se second
#define rtl rt<<1
#define rtr rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define name2str(x) #x
#define fuck(x) cout<<#x" = "<<x<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)+
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("data.txt","r",stdin)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) x&-x
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=a-1;i>=b;--i)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 3e5 + ;
const int maxm = maxn * ;
const int mod = 1e9 + ;
int n, m, a[maxn], b[maxn];
int len, res[maxm], mx; //开大4倍
struct cpx {
long double r, i;
cpx ( long double r = , long double i = ) : r ( r ), i ( i ) {};
cpx operator+ ( const cpx &b ) {
return cpx ( r + b.r, i + b.i );
}
cpx operator- ( const cpx &b ) {
return cpx ( r - b.r, i - b.i );
}
cpx operator* ( const cpx &b ) {
return cpx ( r * b.r - i * b.i, i * b.r + r * b.i );
}
} va[maxm], vb[maxm];
void rader ( cpx F[], int len ) { //len = 2^M,reverse F[i] with F[j] j为i二进制反转
int j = len >> ;
for ( int i = ; i < len - ; ++i ) {
if ( i < j ) swap ( F[i], F[j] ); // reverse
int k = len >> ;
while ( j >= k ) j -= k, k >>= ;
if ( j < k ) j += k;
}
}
void FFT ( cpx F[], int len, int t ) {
rader ( F, len );
for ( int h = ; h <= len; h <<= ) {
cpx wn ( cos ( -t * * pi / h ), sin ( -t * * pi / h ) );
for ( int j = ; j < len; j += h ) {
cpx E ( , ); //旋转因子
for ( int k = j; k < j + h / ; ++k ) {
cpx u = F[k];
cpx v = E * F[k + h / ];
F[k] = u + v;
F[k + h / ] = u - v;
E = E * wn;
}
}
}
if ( t == - ) //IDFT
for ( int i = ; i < len; ++i ) F[i].r /= len;
}
void Conv ( cpx a[], cpx b[], int len ) { //求卷积
FFT ( a, len, );
FFT ( b, len, );
for ( int i = ; i < len; ++i ) a[i] = a[i] * b[i];
FFT ( a, len, - );
}
void gao () {
len = ;
mx = n + m;
while ( len <= mx ) len <<= ; //mx为卷积后最大下标
for ( int i = ; i < len; i++ ) va[i].r = va[i].i = vb[i].r = vb[i].i = ;
for ( int i = ; i < n; i++ ) va[i].r = a[i]; //根据题目要求改写
for ( int i = ; i < m; i++ ) vb[i].r = b[i]; //根据题目要求改写
Conv ( va, vb, len );
for ( int i = ; i < len; ++i ) res[i] += ( LL ) floor ( va[i].r + 0.5 );
} int main() {
// FIN;
sff ( n, m );
int suma = , sumb = , sum1 = , sum2 = ;
for ( int i = ; i < n ; i++ ) sf ( a[n - - i] ), suma += a[n - - i], sum1 += a[n - - i] * a[n - - i];
for ( int i = ; i < n ; i++ ) sf ( b[i] ), b[n + i] = b[i], sumb += b[i], sum2 += b[i] * b[i];
m = * n;
gao();
LL cnt = , ans = 0x3f3f3f3f3f3fLL;
for ( int i = n ; i < * n ; i++ ) cnt = max ( cnt, 1LL * res[i] );
//fuck ( cnt );
for ( int i = -m ; i <= m ; i++ )
ans = min ( ans, 1LL * i * i * n + 1LL * * i * ( suma - sumb ) + sum1 + sum2 - 1LL * * cnt );
printf ( "%lld\n", ans );
return ;
}