牛客国庆集训派对Day1 I Steins;Gate（FFT+原根）

派对Day1 I Steins;Gate（FFT+原根）"/>

牛客国庆集训派对Day1 I Steins;Gate（FFT+原根）

题目链接：

思路：求出p的原根，那么乘法就变成了加法，FFT跑一下，下标在稍微转换一下就好了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll n,p,g,id[200008],a[200008],m,ans[500008];
bool as(ll x)
{ll tmp=1;FOR(i,1,p-2){tmp=tmp*x%p;if(tmp==1) return 0;}return 1;
}
struct CP{double x,y;CP(){} CP(double a,double b):x(a),y(b){}CP operator+(const CP&r) const{return CP(x+r.x,y+r.y);}CP operator-(const CP&r) const{return CP(x-r.x,y-r.y);}CP operator*(const CP&r) const{return CP(x*r.x-y*r.y,x*r.y+y*r.x);}
}b[500008],t;
inline void Swap(CP&a,CP&b) {t=a;a=b;b=t;}
inline void fft(CP*a,int f,int n)
{int i,j,k;for(i=j=0;i<n;i++){if(i>j) Swap(a[i],a[j]);for(k=n>>1;(j^=k)<k;k>>=1);}for(int i=1;i<n;i<<=1){CP wn(cos(pi/i),f*sin(pi/i));for(int j=0;j<n;j+=i<<1){CP w(1,0);for(int k=0;k<i;k++,w=w*wn){CP x=a[j+k],y=w*a[i+j+k];a[j+k]=x+y;a[i+j+k]=x-y;}}}if(f==-1) FOR(i,0,n) a[i].x/=n;
}
int main()
{cin.tie(0);cout.tie(0);cin>>n>>p;for(ll i=2;;i++) if(as(i)) {g=i;break;}ll tmp=1,qw=0,er;for(int i=1;i<p-1;i++){tmp=tmp*g%p;id[tmp]=i;}FOR(i,1,n){sl(a[i]);if(!(a[i]%p)) qw++;else b[id[a[i]%p]].x+=1.0;}er=m=p-1;for(m=er+m,er=1;er<=m;er<<=1);fft(b,1,er);FOR(i,0,er) b[i]=b[i]*b[i];fft(b,-1,er);for(int i=0;i<2*p;i++) ans[i%(p-1)]+=(ll)(b[i].x+0.2);//cout<<ans[id[1]]<<endl;for(int i=1;i<=n;i++){if(a[i]>=p) puts("0");else if (a[i]==0) printf("%lld\n",2*qw*n-qw*qw);else printf("%lld\n",ans[id[a[i]]]);}return 0;
}