XJTUOJ wmq的A×B Problem FFT

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 333333+10, M = 1e3+20,inf = 2e9,mod = 469762049;

int MOD;
inline int mul(int a, int b){
    return (long long)a * b % MOD;
}
int power(int a, int b){
    int ret = 1;
    for (int t = a; b; b >>= 1){
        if (b & 1)ret = mul(ret, t);
        t = mul(t, t);
    }
    return ret;
}
int cal_root(int mod)
{
    int factor[20], num = 0, s = mod - 1;
    MOD = mod--;
    for (int i = 2; i * i <= s; i++){
        if (s % i == 0){
            factor[num++] = i;
            while (s % i == 0)s /= i;
        }
    }
    if (s != 1)factor[num++] = s;
    for (int i = 2;; i++){
        int j = 0;
        for (; j < num && power(i, mod / factor[j]) != 1; j++);
        if (j == num)return i;
    }
}
struct Complex {
    long double r , i ;
    Complex () {}
    Complex ( double r , double i ) : r ( r ) , i ( i ) {}
    Complex operator + ( const Complex& t ) const {
        return Complex ( r + t.r , i + t.i ) ;
    }
    Complex operator - ( const Complex& t ) const {
        return Complex ( r - t.r , i - t.i ) ;
    }
    Complex operator * ( const Complex& t ) const {
        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;
    }
} ;

void FFT ( Complex y[] , int n , int rev ) {
    for ( int i = 1 , j , t , k ; i < n ; ++ i ) {
        for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;
        if ( i < j ) swap ( y[i] , y[j] ) ;
    }
    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {
        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;
        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {
            for ( int i = k ; i < n ; i += s ) {
                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;
                y[i] = y[i] + t ;
            }
        }
    }
    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;
}

Complex s[N];
int T,n,m,x,num[N];
LL ans[N],mo[N],fmo[N];
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        int G = cal_root(m);
        for(LL i = 0, t = 1; i < m-1; ++i,t = t*G%m)
            mo[i] = t,fmo[t] = i;
        memset(num,0,sizeof(num));
        LL cnt0 = 0;
        for(int i = 0; i <= m; ++i) ans[i] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d",&x);
            x%=m;
            if(x == 0) cnt0++;
            else {
                num[fmo[x]]++;
            }
        }
        int n1 = 1;
        for(n1=1;n1<=(2*m-2);n1<<=1);
        for(int i = 0; i < m; ++i) s[i] = Complex(num[i],0);
        for(int i = m; i < n1; ++i) s[i] = Complex(0,0);
        FFT(s,n1,1);
        for(int i = 0; i < n1; ++i) s[i] = s[i]*s[i];
        FFT(s,n1,-1);
        printf("%lld\n",(LL)cnt0*(n-cnt0)+(LL)cnt0*(cnt0-1)/2);
        for(int i = 0; i <= 2*m-2; ++i) {
            LL now = (LL)(s[i].r+0.5);
            if(i%2==0) now -= num[i/2];
            now/=2;
            ans[mo[i%(m-1)]] += now;
        }
        for(int i = 1; i < m; ++i) printf("%lld\n",ans[i]);
    }
    return 0;
}