hihocoder1388 Periodic Signal

来源：互联网发布：杭州小知科技知乎编辑：程序博客网时间：2024/05/01 12:56
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;}
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