hihocode #1388 : Periodic Signal NTT

#1388 : Periodic Signal

描述

Profess X is an expert in signal processing. He has a device which can send a particular 1 second signal repeatedly. The signal is A0 ... An-1 under n Hz sampling.

One day, the device fell on the ground accidentally. Profess X wanted to check whether the device can still work properly. So he ran another n Hz sampling to the fallen device and got B0 ... Bn-1.

To compare two periodic signals, Profess X define the DIFFERENCE of signal A and B as follow:

You may assume that two signals are the same if their DIFFERENCE is small enough. 
Profess X is too busy to calculate this value. So the calculation is on you.

输入

The first line contains a single integer T, indicating the number of test cases.

In each test case, the first line contains an integer n. The second line contains n integers, A0 ... An-1. The third line contains n integers, B0 ... Bn-1.

T≤40 including several small test cases and no more than 4 large test cases.

For small test cases, 0<n≤6⋅103.

For large test cases, 0<n≤6⋅104.

For all test cases, 0≤Ai,Bi<220.

输出

For each test case, print the answer in a single line.

样例输入

2
9
3 0 1 4 1 5 9 2 6
5 3 5 8 9 7 9 3 2
5
1 2 3 4 5
2 3 4 5 1

样例输出

80
0

EmacsNormalVim

 题解：

　　化简式子

　　也就是求

后面这个只需将B数组倒置，进行卷积，出来就是一段连续的位置是k = 0……n-1,所有情况

　　代码来自huyifan

#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define MAXN 300000
typedef long long LL;
const long long P=50000000001507329LL;
const int G=3;

LL mul(LL x,LL y){
    return (x*y-(LL)(x/(long double)P*y+1e-3)*P+P)%P;
}
LL qpow(LL x,LL k,LL p){
    LL ret=1;
    while(k){
        if(k&1) ret=mul(ret,x);
        k>>=1;
        x=mul(x,x);
    }
    return ret;
}

LL wn[25];
void getwn(){
    for(int i=1; i<=18; ++i){
        int t=1<<i;
        wn[i]=qpow(G,(P-1)/t,P);
    }
}

int len;
void NTT(LL y[],int op){
    for(int i=1,j=len>>1,k; i<len-1; ++i){
        if(i<j) swap(y[i],y[j]);
        k=len>>1;
        while(j>=k){
            j-=k;
            k>>=1;
        }
        if(j<k) j+=k;
    }
    int id=0;
    for(int h=2; h<=len; h<<=1) {
        ++id;
        for(int i=0; i<len; i+=h){
            LL w=1;
            for(int j=i; j<i+(h>>1); ++j){
                LL u=y[j],t=mul(y[j+h/2],w);
                y[j]=u+t;
                if(y[j]>=P) y[j]-=P;
                y[j+h/2]=u-t+P;
                if(y[j+h/2]>=P) y[j+h/2]-=P;
                w=mul(w,wn[id]);
            }
        }
    }
    if(op==-1){
        for(int i=1; i<len/2; ++i) swap(y[i],y[len-i]);
        LL inv=qpow(len,P-2,P);
        for(int i=0; i<len; ++i) y[i]=mul(y[i],inv);
    }
}
void Convolution(LL A[],LL B[],int n){
    for(len=1; len<(n<<1); len<<=1);
    for(int i=n; i<len; ++i){
        A[i]=B[i]=0;
    }

    NTT(A,1); NTT(B,1);
    for(int i=0; i<len; ++i){
        A[i]=mul(A[i],B[i]);
    }
    NTT(A,-1);
}
int t,nn,m;
LL a[MAXN],b[MAXN];
LL ans,MAX;
int main()
{

    getwn();
    scanf("%d",&t);

    while(t--)
    {

        MAX=0;

        ans=0;

        scanf("%d",&nn);

        for(int i=0;i<nn;i++)
        {
            scanf("%lld",&a[i]);
            ans+=a[i]*a[i];
        }

        for(int i=0;i<nn;i++)
        {
            scanf("%lld",&b[nn - i - 1]);
            ans+=b[nn - i - 1]*b[nn - i - 1];
        }

        for(int i=0;i<nn;i++)
        {
            a[i+nn]=a[i];
            b[i+nn]=0;
        }

        Convolution(a,b,2*nn);

        for(int i=nn;i<2*nn;i++)
        {
            MAX=max(MAX,a[i]);
        }

        printf("%lld\n",ans-2*MAX);

    }

    return 0;
}