FFT 就可以了 比赛时候没时间做了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int MAXN = 6e4+5;
int A[MAXN<<2], B[MAXN<<2], C[MAXN<<2];
struct FFTSOLVE {
int pos[MAXN<<2];
struct comp {
double r , i ;
comp ( double _r = 0 , double _i = 0 ) : r ( _r ) , i ( _i ) {}
comp operator + ( const comp& x ) {
return comp ( r + x.r , i + x.i ) ;
}
comp operator - ( const comp& x ) {
return comp ( r - x.r , i - x.i ) ;
}
comp operator * ( const comp& x ) {
return comp ( r * x.r - i * x.i , i * x.r + r * x.i ) ;
}
comp conj () {
return comp ( r , -i ) ;
}
} A[MAXN<<2] , B[MAXN<<2] ;
const double pi = acos ( -1.0 ) ;
void FFT ( comp a[] , int n , int t ) {
for ( int i = 1 ; i < n ; ++ i ) if ( pos[i] > i ) swap ( a[i] , a[pos[i]] ) ;
for ( int d = 0 ; ( 1 << d ) < n ; ++ d ) {
int m = 1 << d , m2 = m << 1 ;
double o = pi * 2 / m2 * t ;
comp _w ( cos ( o ) , sin ( o ) ) ;
for ( int i = 0 ; i < n ; i += m2 ) {
comp w ( 1 , 0 ) ;
for ( int j = 0 ; j < m ; ++ j ) {
comp& A = a[i + j + m] , &B = a[i + j] , t = w * A ;
A = B - t ;
B = B + t ;
w = w * _w ;
}
}
}
if ( t == -1 ) for ( int i = 0 ; i < n ; ++ i ) a[i].r /= n ;
}
void mul ( int *a , int *b , int *c ,int k) {
int i , j ;
for ( i = 0 ; i < k ; ++ i ) A[i] = comp ( a[i] , b[i] ) ;
j = __builtin_ctz ( k ) - 1 ;
for ( int i = 0 ; i < k ; ++ i ) {
pos[i] = pos[i >> 1] >> 1 | ( ( i & 1 ) << j ) ;
}
FFT ( A , k , 1 ) ;
for ( int i = 0 ; i < k ; ++ i ) {
j = ( k - i ) & ( k - 1 ) ;
B[i] = ( A[i] * A[i] - ( A[j] * A[j] ).conj () ) * comp ( 0 , -0.25 ) ;
}
FFT ( B , k , -1 ) ;
for ( int i = 0 ; i < k ; ++ i ) {
c[i] = ( long long ) ( B[i].r + 0.5 );
}
}
}boy;
int N;
int s1[MAXN], s2[MAXN];
int main(){
int T; scanf("%d",&T);
while(T--) {
ll ans = 0;
memset(A,0,sizeof(A));
memset(B,0,sizeof(B));
scanf("%d",&N);
for(int i = 0; i < N; ++i) {
scanf("%d",&s1[i]);
}
for(int i = 0; i < N; ++i) {
scanf("%d",&s2[i]);
}
int len = 1; while(len < N*2) len <<= 1;
for(int i = 0; i < N; ++i) {
A[i] = s1[N-i-1]; B[i] = s2[i];
}
for(int i = N; i < 2*N; ++i) {
A[i] = 0; B[i] = s2[i-N];
}
boy.mul(A,B,C,len);
double an = -1; double eps = 1e-8; int pos;
for(int i = 0; i < N; ++i) {
if(an+eps < boy.B[i+N-1].r) {
an = boy.B[i+N-1].r; pos = i;
}
}
for(int i = 0; i < N; ++i) {
ans += 1ll*(s1[i]-s2[(i+pos)%N]) * (s1[i]-s2[(i+pos)%N]);
}
printf("%lld\n",ans);
}
return 0;
}