#1388 : Periodic Signal

/*hihocoder 1388 fft循环矩阵problem:给你两个数组,求差值平方和的最小值.  (a[i] - b[(i+k)%n])^solve:可以转换成 (a[i]*a[i]-2*a[i]*b[j]+b[j]*b[j]). 也就成了求a[i]*b[j]的最大值. 然后就没什么想法了- -参考：http://blog.csdn.net/VictorZC8/article/details/52655099说的是可以转换成两个数组相乘. 这样的话fft就能解决了. double型的话会有精度问题结果发现别人是先求fft求完,比较出k的价值. 然后计算 ( 好机制Orz )hhh-2016-09-25 16:53:25*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <map>#include <math.h>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define key_val ch[ch[root][1]][0]using namespace std;const int maxn = 1e6 + 1000;const int inf = 0x3f3f3f3f;const ll mod = 1e9 + 7;const double eps = 1e-7;template<class T> void read(T&num){    char CH;    bool F=false;    for(CH=getchar(); CH<‘0‘||CH>‘9‘; F= CH==‘-‘,CH=getchar());    for(num=0; CH>=‘0‘&&CH<=‘9‘; num=num*10+CH-‘0‘,CH=getchar());    F && (num=-num);}int stk[70], tp;template<class T> inline void print(T p){    if(!p)    {        puts("0");        return;    }    while(p) stk[++ tp] = p%10, p/=10;    while(tp) putchar(stk[tp--] + ‘0‘);    putchar(‘\n‘);}const double PI = acos(-1.0);struct Complex{    double x,y;    Complex(double _x = 0.0,double _y = 0.0)    {        x = _x;        y = _y;    }    Complex operator-(const Complex &b)const    {        return Complex(x-b.x,y-b.y);    }    Complex operator+(const Complex &b)const    {        return Complex(x+b.x,y+b.y);    }    Complex operator*(const Complex &b)const    {        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);    }};void change(Complex y[],int len){    int i,j,k;    for(i = 1,j = len/2; i < len-1; i++)    {        if(i < j) swap(y[i],y[j]);        k = len/2;        while(j >= k)        {            j-=k;            k/=2;        }        if(j < k) j+=k;    }}void fft(Complex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));        for(int j = 0; j < len; j+=h)        {            Complex w(1,0);            for(int k = j; k < j+h/2; k++)            {                Complex u = y[k];                Complex t = w*y[k+h/2];                y[k] = u+ t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)    {        for(int i = 0; i < len; i++)            y[i].x /= len;    }}Complex a[maxn];Complex b[maxn];ll x[maxn],y[maxn];int n;void File(){    freopen("in.txt","r",stdin);}int main(){    int T;//    File();    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        int len = 1;        while(len <= (n*2 -1)) len <<= 1;        for(int i = 0;i < n;i++)        {            scanf("%lld",&x[i]);            a[i] = x[i];        }        for(int i = n;i < len;i++)            a[i] = 0;        for(int i = 0;i < n;i++)        {            scanf("%lld",&y[i]);        }        for(int i = 0;i < n;i++) b[i] = y[n-1-i];        for(int i = 0;i < n-1;i++) b[i+n] = y[n-1-i];        for(int i = 2*n-1;i < len;i ++)            b[i] = 0;        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);        ll Max = -1;        int pos = 0;        for(int i = n-1,j = 0;j <= n;j++ )        {            if( (ll)(a[i+j].x + 0.5) > Max)            {                pos = j;                Max = (ll)(a[i+j].x + 0.5);            }        }        ll ans = 0;        pos = (n-pos) % n;//        cout << pos<<endl;        for(int i = 0;i < n;i++)        {            ans += (x[i] - y[(i + pos) % n])*(x[i] - y[(i + pos) % n]);        }        print(ans);    }    return 0;}