2 条题解
-
1

#include <bits/stdc++.h> #define N 263000 #define lg2(x) (31 - __builtin_clz(x)) typedef long long ll; const ll mod = 998244353, root = 31; int n; int a[N], b[N], c[N]; ll fact[N], finv[N]; ll PowerMod(ll a, int n, ll c = 1) {for (; n; n >>= 1, a = a * a % mod) if (n & 1) c = c * a % mod; return c;} namespace Poly { ll iv; int l, n; int x[N], y[N], rev[N]; void NTT_init(int len) { n = 1 << (l = len); iv = PowerMod(n, mod - 2); ll g = PowerMod(root, 1 << 23 - l); x[0] = 1; rev[0] = 0; for (int i = 1; i < n; ++i){ x[i] = (ll)x[i - 1] * g % mod; rev[i] = (i & 1 ? rev[i - 1] | 1 << (l - 1) : rev[i >> 1] >> 1); } } void DNTT(int *d, int *t) { int i, *j, *k, len = 1, delta = n, R; for (i = 0; i < n; ++i) t[i] = d[rev[i]]; for (i = 0; i < l; ++i) { delta >>= 1; for (k = x, j = y; j < y + len; k += delta, ++j) *j = *k; for (j = t; j < t + n; j += len << 1) for (k = j; k < j + len; ++k) { R = (ll)y[k - j] * k[len] % mod; k[len] = (*k - R < 0 ? *k - R + mod : *k - R); *k = (*k + R >= mod ? *k + R - mod : *k + R); } len <<= 1; } } void Mul(int *a, int *b, int *c, int deg){ int i; NTT_init(lg2(deg) + 1); DNTT(a, c); DNTT(b, a); for(i = 0; i < n; ++i) a[i] = (ll)a[i] * c[i] % mod; DNTT(a, c); std::reverse(c + 1, c + n); for(i = 0; i < n; i++) c[i] = (ll)c[i] * iv % mod; } } int main() { int i; ll r, coe = 1, ans = 0; scanf("%d", &n); for (*fact = i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i % mod; finv[n] = PowerMod(fact[n], mod - 2); for (i = n; i; --i) finv[i - 1] = finv[i] * i % mod; for (i = 0; i <= n; ++i) a[i] = i & 1 ? mod - finv[i] : finv[i]; b[0] = 1; b[1] = n + 1; for (i = 2; i <= n; ++i) { r = finv[i] * finv[i - 1] % mod * fact[i - 2] % mod; b[i] = (PowerMod(i, n + 1, r) + mod - r) % mod; } Poly::Mul(a, b, c, n << 1); for (i = 0; i <= n; ++i) { ans = (ans + c[i] * coe) % mod; coe = coe * 2 * (i + 1) % mod; } printf("%lld\n", ans); return 0; } -
0
#include<bits/stdc++.h> using namespace std; #define int long long #define fu(i,j,k) for(int i=j;i<=k;i++) #define fd(i,j,k) for(int i=j;i>=k;i--) const int N=4e5+10,P=998244353; int qpow(int a,int b){int ans=1;for(;b;b>>=1,a=a*a%P)if(b&1)ans=ans*a%P;return ans;} int a[N],b[N],c[N],fac[N],d2[N]; void ntt(int s[],int n,int x) { if(n==1)return; int s1[n/2],s2[n/2]; fu(i,0,n/2-1)s1[i]=s[i*2],s2[i]=s[i*2+1]; ntt(s1,n/2,x*x%P),ntt(s2,n/2,x*x%P); for(int i=0,xi=1;i<n/2;i++,xi=xi*x%P) { s[i]=(s1[i]+s2[i]*xi)%P; s[i+(n/2)]=((s1[i]-s2[i]*xi)%P+P)%P; } } signed main() { int n;cin>>n; fac[0]=1;fu(i,1,n)fac[i]=fac[i-1]*i%P; d2[0]=1;fu(i,1,n)d2[i]=d2[i-1]*2%P; fu(i,0,n)a[i]=(((i%2==0)?1:-1)*qpow(fac[i],P-2)%P+P)%P; b[0]=1;b[1]=n+1;fu(i,2,n)b[i]=(qpow(i,n+1)-1)*qpow((i-1)*fac[i]%P,P-2)%P; int D=1;while(D<n*2-1)D<<=1; int inv=qpow(D,P-2),x=qpow(3,(P-1)/D); ntt(a,D,x);ntt(b,D,x); fu(i,0,D-1)c[i]=a[i]*b[i]%P; int inv1=qpow(x,P-2); ntt(c,D,inv1); fu(i,0,n*2-2)c[i]=c[i]*inv%P; int ans=0; fu(i,0,n)ans=(ans+c[i]*fac[i]%P*d2[i]%P)%P; cout<<ans; return 0; }
- 1
信息
- ID
- 6220
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 8
- 标签
- 递交数
- 20
- 已通过
- 7
- 上传者