BZOJ 4827 循环卷积

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("******
")
#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 + 7;
const int maxm = maxn * 4;
const int mod = 1e9 + 7;
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 = 0, long double i = 0 ) : 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 >> 1;
    for ( int i = 1; i < len - 1; ++i ) {
        if ( i < j ) swap ( F[i], F[j] ); // reverse
        int k = len >> 1;
        while ( j >= k ) j -= k, k >>= 1;
        if ( j < k ) j += k;
    }
}
void FFT ( cpx F[], int len, int t ) {
    rader ( F, len );
    for ( int h = 2; h <= len; h <<= 1 ) {
        cpx wn ( cos ( -t * 2 * pi / h ), sin ( -t * 2 * pi / h ) );
        for ( int j = 0; j < len; j += h ) {
            cpx E ( 1, 0 ); //旋转因子
            for ( int k = j; k < j + h / 2; ++k ) {
                cpx u = F[k];
                cpx v = E * F[k + h / 2];
                F[k] = u + v;
                F[k + h / 2] = u - v;
                E = E * wn;
            }
        }
    }
    if ( t == -1 ) //IDFT
        for ( int i = 0; i < len; ++i ) F[i].r /= len;
}
void Conv ( cpx a[], cpx b[], int len ) { //求卷积
    FFT ( a, len, 1 );
    FFT ( b, len, 1 );
    for ( int i = 0; i < len; ++i ) a[i] = a[i] * b[i];
    FFT ( a, len, -1 );
}
void gao () {
    len = 1;
    mx = n + m;
    while ( len <= mx ) len <<= 1; //mx为卷积后最大下标
    for ( int i = 0; i < len; i++ ) va[i].r = va[i].i = vb[i].r = vb[i].i = 0;
    for ( int i = 0; i < n; i++ ) va[i].r = a[i]; //根据题目要求改写
    for ( int i = 0; i < m; i++ ) vb[i].r = b[i]; //根据题目要求改写
    Conv ( va, vb, len );
    for ( int i = 0; i < len; ++i ) res[i] += ( LL ) floor ( va[i].r + 0.5 );
}

int main() {
   // FIN;
    sff ( n, m );
    int suma = 0, sumb = 0, sum1 = 0, sum2 = 0;
    for ( int i = 0 ; i < n ; i++ ) sf ( a[n - 1 - i] ), suma += a[n - 1 - i], sum1 += a[n - 1 - i] * a[n - 1 - i];
    for ( int i = 0 ; i < n ; i++ ) sf ( b[i] ), b[n + i] = b[i], sumb += b[i], sum2 += b[i] * b[i];
    m = 2 * n;
    gao();
    LL cnt = 0, ans = 0x3f3f3f3f3f3fLL;
    for ( int i = n ; i < 2 * 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 * 2 * i * ( suma - sumb ) + sum1 + sum2 - 1LL * 2 * cnt );
    printf ( "%lld
", ans );
    return 0;
}