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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#197762 | #6349. Is This FFT? | zhouhuanyi | TL | 0ms | 22276kb | C++23 | 3.6kb | 2023-10-02 19:36:16 | 2023-10-02 19:36:16 |
Judging History
answer
#include<iostream>
#include<cstdio>
#define N 250
#define M 62500
#define K 65536
#define Q 16
using u32 = unsigned int;
using u64 = unsigned long long;
using u128 = __uint128_t;
using namespace std;
struct Barrett {u64 b, m;Barrett() : b(), m() {}Barrett(u64 _b) : b(_b), m(-1ULL / _b) {}u32 reduce(u64 x) {u64 q = (u64)((u128(m) * x) >> 64), r = x - q * b;return r - b * (r >= b);}} BA;
u32 mult(u32 x, u32 y) {return BA.reduce((u64)x * y);}
int read()
{
char c=0;
int sum=0;
while (c<'0'||c>'9') c=getchar();
while ('0'<=c&&c<='9') sum=sum*10+c-'0',c=getchar();
return sum;
}
int n,g,length,ans,st[N+1],leng,rev[K+1],S[N+1],F[K+1],fac[M+1],invfac[M+1],inv[M+1],dp[N+1][M+1],DP[N+1][K+1],wn[Q+1][K+1],wn2[Q+1][K+1],num[K+1],mod;
void Adder(int &x,int d)
{
x+=d;
if (x>=mod) x-=mod;
return;
}
void Adder2(int &x,int d)
{
x+=d;
if (x<0) x+=mod;
return;
}
int MD(int x)
{
return x>=mod?x-mod:x;
}
int MD2(int x)
{
return x<0?x+mod:x;
}
int fast_pow(int a,int b)
{
int res=1,mul=a;
while (b)
{
if (b&1) res=mult(res,mul);
mul=mult(mul,mul),b>>=1;
}
return res;
}
int C(int x,int y)
{
if (x<y) return 0;
return mult(mult(fac[x],invfac[y]),invfac[x-y]);
}
void NTT(int limit,int *s,int type)
{
int s1,s2;
for (int i=0;i<limit;++i)
if (rev[i]>i)
swap(s[i],s[rev[i]]);
if (type==1)
{
for (int i=2;i<=limit;i<<=1)
for (int j=0;j+i-1<limit;j+=i)
for (int k=j;k<j+(i>>1);++k)
s1=s[k],s2=mult(s[k+(i>>1)],wn[num[i]][k-j]),s[k]=MD(s1+s2),s[k+(i>>1)]=MD2(s1-s2);
}
else
{
for (int i=2;i<=limit;i<<=1)
for (int j=0;j+i-1<limit;j+=i)
for (int k=j;k<j+(i>>1);++k)
s1=s[k],s2=mult(s[k+(i>>1)],wn2[num[i]][k-j]),s[k]=MD(s1+s2),s[k+(i>>1)]=MD2(s1-s2);
s1=fast_pow(limit,mod-2);
for (int i=0;i<=limit;++i) s[i]=mult(s[i],s1);
}
return;
}
int main()
{
int x,w,limit=1;
bool op;
n=read(),mod=read(),x=mod-1,fac[0]=1,BA=Barrett(mod);
for (int i=1;i<=M;++i) fac[i]=mult(fac[i-1],i);
invfac[M]=fast_pow(fac[M],mod-2);
for (int i=M-1;i>=0;--i) invfac[i]=mult(invfac[i+1],i+1);
for (int i=1;i<=M;++i) inv[i]=mult(fac[i-1],invfac[i]);
for (int i=2;i*i<=(mod-1);++i)
if (x%i==0)
{
while (x%i==0) x/=i;
st[++leng]=i;
}
if (x!=1) st[++leng]=x;
while (1)
{
g++,op=1;
for (int i=1;i<=leng;++i)
if (fast_pow(g,(mod-1)/st[i])==1)
op=0;
if (op) break;
}
for (int i=2;i<=K;i<<=1)
{
num[i]=++length,w=fast_pow(g,(mod-1)/i);
for (int j=0,res=1;j<(i>>1);++j,res=mult(res,w)) wn[num[i]][j]=res;
w=fast_pow(g,(mod-1)/i*(i-1));
for (int j=0,res=1;j<(i>>1);++j,res=mult(res,w)) wn2[num[i]][j]=res;
}
for (int i=1;i<=n;++i) S[i]=S[i-1]+i-1;
while (limit<=S[n]) limit<<=1;
for (int i=1;i<limit;++i) rev[i]=(rev[i>>1]>>1)|((i&1)?limit>>1:0);
dp[1][0]=DP[1][0]=1,NTT(limit,DP[1],1);
for (int i=2;i<=n;++i)
for (int j=0;j<=limit;++j)
DP[i][j]=DP[1][j];
for (int i=2;i<=n;++i)
{
for (int j=1;j<=i-1;++j)
{
for (int k=0;k<=limit;++k) F[k]=0;
for (int k=1;k<=j*(i-j);++k) F[k]=(k&1)?C(j*(i-j)-1,k-1):MD2(-C(j*(i-j)-1,k-1));
NTT(limit,F,1);
for (int k=0;k<=limit;++k) Adder(DP[i][k],mult(mult(DP[j][k],DP[i-j][k]),F[k]));
}
NTT(limit,DP[i],-1);
for (int j=1;j<=S[i];++j) dp[i][j]=1ll*DP[i][j]*inv[j]%mod;
for (int j=0;j<=limit;++j) DP[i][j]=0;
for (int j=1;j<=S[i];++j) DP[i][j]=dp[i][j];
NTT(limit,DP[i],1);
}
for (int i=2;i<=n;++i)
{
ans=0;
for (int j=1;j<=S[i];++j) Adder(ans,dp[i][j]);
printf("%lld\n",1ll*ans*fac[i]%mod*inv[2]%mod);
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 22276kb
input:
10 998244353
output:
1 1 532396989 328786831 443364983 567813846 34567523 466373946 474334062
result:
ok 9 numbers
Test #2:
score: -100
Time Limit Exceeded
input:
250 998244353