原根求解算法：
获取一个数\(N\)的原根\(root\)的算法
#include<bits/stdc++.h>
#define ll long long
#define IL inline
#define RG register
using namespace std;
ll prm[1000],tot,N,root;
ll Power(ll bs,ll js,ll MOD){
ll S = 1,T = bs;
while(js){
if(js&1)S = S*T%MOD;
T = T*T%MOD;
js >>= 1;
} return S;
}
IL ll GetRoot(RG ll n){
RG ll tmp = n - 1 , tot = 0;
for(RG ll i = 2; i <= sqrt(tmp); i ++){
if(tmp%i==0){
prm[++tot] = i;
while(tmp%i==0)tmp /= i;
}
}
if(tmp != 1)prm[++tot] = tmp; //质因数分解
for(RG ll g = 2; g <= n-1; g ++){
bool flag = 1;
for(RG int i = 1; i <= tot; i ++){ //检测是否符合条件
if(Power(g,(n-1)/prm[i],n) == 1)
{ flag = 0; break; }
}
if(flag)return g;
}return 0; //无解
}
int main(){
cin >> N;
root = GetRoot(N);
cout<<root<<endl;
return 0;
}
快速数论变换算法：
计算多项式\(f_1*f_2\)在模\(P\) (\(P\)为质数) 意义下的卷积。
讲真的，只要把\(FFT\)的单位复数根换成原根就行了。
注意要提前用上面的算法把模数的原根算出来。
#define mod 998244353 //使用NTT需要保证模数mod 为质数
const ll pr = 3;
//3是998244353的原根，在比赛中请用上面那个算法提前算出....
ll f1[_],f2[_],U,V;
ll wn[50],R[_],N,M,n,m,l,ans[_];
IL ll Power(RG ll bs,RG ll js){
RG ll S = 1 , T = bs;
while(js){if(js&1)S=S*T%mod; T=T*T%mod; js>>=1;}
return S;
}
IL void GetWn(){
//需要计算floor(log n)个原根
for(RG int i = 0; i <= 25; i ++){
RG ll tt = 1<<i;
wn[i] = Power(pr,(mod-1)/tt);
}return;
}
IL void NTT(RG ll P[],RG int opt){
for(RG int i = 0; i < n; i ++)
if(i < R[i]) swap(P[R[i]],P[i]);
for(RG int i = 1,id = 0; i < n; i<<=1){
id ++;
for(RG int j = 0,p = i<<1; j < n; j += p){
RG ll w = 1;
for(RG int k = 0; k < i; k ++,w = w*wn[id]%mod){
U = P[j+k]; V = w*P[j+k+i];
P[j+k] = (U+V)%mod; P[j+k+i] = ((U-V)%mod+mod)%mod;
}
}
}
if(opt == -1){
//caution:反转时是从1开始 for !!!!!
for(RG int i = 1; i < n/2; i ++)swap(P[i],P[n-i]);
RG ll inv = Power(n,mod-2);
for(RG int i = 0; i < n; i ++)P[i] = P[i]%mod*inv%mod;
}return;
}
int main(){
//读入数据：
cin >> N >> M;
for(RG int i = 0; i <= N; i ++)cin >> f1[i];
for(RG int i = 0; i <= M; i ++)cin >> f2[i];
//NTT计算：
m = N+M; l = 0;
for(n = 1; n <= m; n<<=1) ++ l;
for(RG int i = 0; i < n; i ++)
R[i] = (R[i>>1]>>1) | ((i&1)<<(l-1));
GetWn();
NTT(f1,1); NTT(f2,1);
for(RG int i = 0; i < n; i ++)f1[i] = f1[i]*f2[i]%mod;
NTT(f1,-1);
//转移答案：
for(RG int i = 0; i <= m; i ++)ans[i] = f1[i];
for(RG int i = 0; i <= m; i ++)cout<<ans[i]<<" ";
return 0;
}