• 与数论有关的模板


    有一些数论相关的模板,经常是需要用到的。


    下降幂多项式乘法

    $$f(x)=sum_{i=0}^na_ix^{underline{i}}$$

    $$G(x)=sum_{i=0}^na_ifrac{x^i}{i!}$$

    $$F(x)=sum_{i=0}^nf(i)frac{x^i}{i!}$$

    其中$x^{underline{n}}$表示$prod_{i=0}^{n-1}(x-i)$,类似的还有$x^{overline{n}}$表示上升幂$prod_{i=0}^{n-1}(x+i)$。

    即$A(x)$为原多项式,$G(x)$为系数的EGF,$F(x)$为点值的EGF。

    我们发现$x^{underline{n}}$对$F(x)$的贡献为$sum_{i=n}^{infty}frac{i^{underline{n}}}{i!}x^i=sum_{i=n}^{infty}frac{1}{(i-n)!}x^i=x^ne^x$

    所以$F=G*e^x$,即$G=F*e^{-x}$

    点值就可以直接相乘了。

     1 #include<bits/stdc++.h>
     2 #define Rint register int
     3 using namespace std;
     4 typedef long long LL;
     5 const int mod = 998244353, G = 3, Gi = 332748118, N = 800003;
     6 inline int kasumi(int a, int b){
     7     int res = 1;
     8     while(b){
     9         if(b & 1) res = (LL) res * a % mod;
    10         a = (LL) a * a % mod;
    11         b >>= 1;
    12     }
    13     return res;
    14 }
    15 int fac[N], inv[N];
    16 inline void init(int n){
    17     fac[0] = 1;
    18     for(Rint i = 1;i <= n;i ++) fac[i] = (LL) i * fac[i - 1] % mod;
    19     inv[n] = kasumi(fac[n], mod - 2);
    20     for(Rint i = n;i;i --) inv[i - 1] = (LL) i * inv[i] % mod;
    21 }
    22 int rev[N];
    23 inline int calrev(int n){
    24     int limit = 1, L = -1;
    25     while(limit <= n){limit <<= 1; L ++;}
    26     for(Rint i = 0;i < limit;i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    27     return limit;
    28 }
    29 inline void NTT(int *A, int limit, int type){
    30     for(Rint i = 0;i < limit;i ++) if(i < rev[i]) swap(A[i], A[rev[i]]);
    31     for(Rint mid = 1;mid < limit;mid <<= 1){
    32         int Wn = kasumi(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
    33         for(Rint j = 0;j < limit;j += (mid << 1)){
    34             int w = 1;
    35             for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
    36                 int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
    37                 A[j + k] = (x + y) % mod;
    38                 A[j + k + mid] = (x - y + mod) % mod;
    39             }
    40         }
    41     }
    42     if(type == -1){
    43         int inv = kasumi(limit, mod - 2);
    44         for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
    45     }
    46 }
    47 int n, m, A[N], B[N], C[N];
    48 int main(){
    49     scanf("%d%d", &n, &m);
    50     for(Rint i = 0;i <= n;i ++) scanf("%d", A + i);
    51     for(Rint i = 0;i <= m;i ++) scanf("%d", B + i);
    52     init(n + m);
    53     for(Rint i = 0;i <= n + m;i ++) C[i] = inv[i];
    54     int limit = calrev((n + m) << 1);
    55     NTT(A, limit, 1); NTT(B, limit, 1); NTT(C, limit, 1);
    56     for(Rint i = 0;i < limit;i ++){
    57         A[i] = (LL) A[i] * C[i] % mod;
    58         B[i] = (LL) B[i] * C[i] % mod;
    59     }
    60     NTT(A, limit, -1); NTT(B, limit, -1);
    61     for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * B[i] % mod * fac[i] % mod;
    62     memset(C, 0, sizeof C);
    63     for(Rint i = 0;i <= n + m;i ++) C[i] = (i & 1) ? (mod - inv[i]) : inv[i];
    64     NTT(A, limit, 1); NTT(C, limit, 1);
    65     for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * C[i] % mod;
    66     NTT(A, limit, -1);
    67     for(Rint i = 0;i <= n + m;i ++) printf("%d ", A[i]);
    68 }
    Luogu5394

    点值转下降幂多项式

    详见UOJ269【清华集训2016】如何优雅地求和


    通常幂,下降幂,上升幂互相转化

    通常幂转化为其他两个用第二类斯特林数.

    转化为通常幂用第一类斯特林数

    小转大的时候带上$(-1)^{n-i}$的系数。


    第二类斯特林数-行

    $$egin{Bmatrix}n \ mend{Bmatrix}=sum_{i=0}^mfrac{(-1)^i}{i!} imes frac{(m-i)^n}{(m-i)!}$$

     1 #include<bits/stdc++.h>
     2 #define Rint register int
     3 using namespace std;
     4 typedef long long LL;
     5 const int N = 800003, mod = 167772161, g = 3, gi = 55924054;
     6 int n, inv[N], A[N], B[N];
     7 inline int kasumi(int a, int b){
     8     int res = 1;
     9     while(b){
    10         if(b & 1) res = (LL) res * a % mod;
    11         a = (LL) a * a % mod;
    12         b >>= 1;
    13     }
    14     return res;
    15 }
    16 int rev[N];
    17 inline void NTT(int *A, int limit, int type){
    18     for(Rint i = 0;i < limit;i ++)
    19         if(i < rev[i]) swap(A[i], A[rev[i]]);
    20     for(Rint mid = 1;mid < limit;mid <<= 1){
    21         int Wn = kasumi(type == 1 ? g : gi, (mod - 1) / (mid << 1));
    22         for(Rint j = 0;j < limit;j += mid << 1){
    23             int w = 1;
    24             for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
    25                 int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
    26                 A[j + k] = (x + y) % mod;
    27                 A[j + k + mid] = (x - y + mod) % mod;
    28             }
    29         }
    30     }
    31     if(type == -1){
    32         int inv = kasumi(limit, mod - 2);
    33         for(Rint i = 0;i < limit;i ++)
    34             A[i] = (LL) A[i] * inv % mod;
    35     }
    36 }
    37 int main(){
    38     scanf("%d", &n);
    39     inv[1] = 1;
    40     for(Rint i = 2;i <= n;i ++) inv[i] = (LL) (mod - mod / i) * inv[mod % i] % mod;
    41     inv[0] = 1;
    42     for(Rint i = 1;i <= n;i ++) inv[i] = (LL) inv[i] * inv[i - 1] % mod;
    43     for(Rint i = 0;i <= n;i ++){
    44         A[i] = (i & 1) ? -inv[i] : inv[i];
    45         B[i] = (LL) kasumi(i, n) * inv[i] % mod;
    46     }
    47     int limit = 1, L = -1;
    48     while(limit <= (n + 1 << 1)){limit <<= 1; L ++;}
    49     for(Rint i = 0;i < limit;i ++)
    50         rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    51     NTT(A, limit, 1); NTT(B, limit, 1);
    52     for(Rint i = 0;i < limit;i ++)
    53         A[i] = (LL) A[i] * B[i] % mod;
    54     NTT(A, limit, -1);
    55     for(Rint i = 0;i <= n;i ++)
    56         printf("%d ", A[i]);
    57 }
    Luogu5395

    第二类斯特林数-列

    $$S_m(x)=sum_{i=0}^{+infty}x^iegin{Bmatrix}i \ mend{Bmatrix}$$

    $$egin{Bmatrix}n \ mend{Bmatrix}=egin{Bmatrix}n-1 \ m-1end{Bmatrix}+egin{Bmatrix}n-1 \ mend{Bmatrix} imes m$$

    $$S_m(x)=mxS_m(x)+xS_{m-1}(x)$$

    $$S_m(x)=S_{m-1}(x) imes frac{x}{1-mx}$$

    $$S_m(x)=frac{x^m}{prod_{i=1}^m(1-ix)}$$

    分治计算连乘积,$O(nlog^2 n)$。

    $$sum_{i=k}^{+infty}egin{Bmatrix}i \ kend{Bmatrix}frac{x^i}{i!}=frac{(e^x-1)^k}{k!}$$

    这个式子的组合意义就是,左边是把$i$个无区别的元素分成$k$个无序的集合,右边表示每个集合的EGF是$e^x-1$,然后去标号是除以$k!$。

    虽然这个做法是$O(nlog n)$,但是实际上更慢。(exp大常数)

     1 #include<bits/stdc++.h>
     2 #define Rint register int
     3 using namespace std;
     4 typedef long long LL;
     5 const int N = 1 << 19, mod = 167772161, g = 3, gi = 55924054;
     6 int n, m;
     7 inline int kasumi(int a, int b){
     8     int res = 1;
     9     while(b){
    10         if(b & 1) res = (LL) res * a % mod;
    11         a = (LL) a * a % mod;
    12         b >>= 1;
    13     }
    14     return res;
    15 }
    16 int rev[N];
    17 inline void NTT(int *A, int limit, int type){
    18     for(Rint i = 0;i < limit;i ++)
    19         if(i < rev[i]) swap(A[i], A[rev[i]]);
    20     for(Rint mid = 1;mid < limit;mid <<= 1){
    21         int Wn = kasumi(type == 1 ? g : gi, (mod - 1) / (mid << 1));
    22         for(Rint j = 0;j < limit;j += (mid << 1)){
    23             int w = 1;
    24             for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
    25                 int x = A[j + k], y = (LL) A[j + k + mid] * w % mod;
    26                 A[j + k] = (x + y) % mod;
    27                 A[j + k + mid] = (x - y + mod) % mod;
    28             }
    29         }
    30     }
    31     if(type == -1){
    32         int inv = kasumi(limit, mod - 2);
    33         for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
    34     }
    35 }
    36 int ans[N];
    37 inline void poly_inv(int *A, int deg){
    38     static int tmp[N];
    39     if(deg == 1){
    40         ans[0] = kasumi(A[0], mod - 2);
    41         return;
    42     }
    43     poly_inv(A, deg + 1 >> 1);
    44     int limit = 1, L = -1;
    45     while(limit <= (deg << 1)){limit <<= 1; L ++;}
    46     for(Rint i = 0;i < limit;i ++)
    47         rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    48     for(Rint i = 0;i < deg;i ++) tmp[i] = A[i];
    49     for(Rint i = deg;i < limit;i ++) tmp[i] = 0;
    50     NTT(tmp, limit, 1); NTT(ans, limit, 1);
    51     for(Rint i = 0;i < limit;i ++)
    52         ans[i] = (2 - (LL) tmp[i] * ans[i] % mod + mod) % mod * ans[i] % mod;
    53     NTT(ans, limit, -1);
    54     for(Rint i = deg;i < limit;i ++) ans[i] = 0;
    55 }
    56 int A[20][N];
    57 inline void calc(int dep, int L, int R){
    58     if(L == R){
    59         A[dep][0] = 1; A[dep][1] = mod - L;
    60         return;
    61     }
    62     int mid = L + R >> 1;
    63     calc(dep + 1, L, mid);
    64     for(Rint i = 0;i <= mid - L + 1;i ++) A[dep][i] = A[dep + 1][i];
    65     calc(dep + 1, mid + 1, R);
    66     int limit = 1, l = -1;
    67     while(limit <= (R - L + 1)){limit <<= 1; l ++;}
    68     for(Rint i = 0;i < limit;i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l);
    69     for(Rint i = mid - L + 2;i < limit;i ++) A[dep][i] = 0;
    70     for(Rint i = R - mid + 1;i < limit;i ++) A[dep + 1][i] = 0;
    71     NTT(A[dep], limit, 1); NTT(A[dep + 1], limit, 1);
    72     for(Rint i = 0;i < limit;i ++) A[dep][i] = (LL) A[dep][i] * A[dep + 1][i] % mod;
    73     NTT(A[dep], limit, -1);
    74 }
    75 int main(){
    76 //    freopen("my.out", "w", stdout);
    77     scanf("%d%d", &n, &m);
    78     if(n < m){
    79         for(Rint i = 0;i <= n;i ++) printf("0 ");
    80         return 0;
    81     }
    82     calc(0, 1, m);
    83 //    for(Rint i = 0;i <= m;i ++) printf("%d ", A[0][i]);
    84 //    putchar('
    ');
    85     poly_inv(A[0], n - m + 1);
    86     for(Rint i = 0;i < m;i ++) printf("0 ");
    87     for(Rint i = 0;i <= n - m;i ++) printf("%d ", ans[i]);
    88 }
    Luogu5396

    第一类斯特林数-列

    $$(x+1)^k=sum_{i=0}^kx^ifrac{k^{underline{i}}}{i!}$$

    $$=sum_{i=0}^kfrac{x^i}{i!}sum_{j=0}^i(-1)^{i-j}egin{bmatrix}i \ jend{bmatrix}k^j$$

    $$=sum_{i=0}^nk^isum_{j=i}^k(-1)^{j-i}egin{bmatrix}j \ iend{bmatrix}frac{x^j}{j!}$$

    用另一种方法展开可以得到:

    $$(x+1)^k=exp(kln(x+1))=sum_{i=0}^{+infty}k^ifrac{ln(x+1)^i}{i!}$$

    所以

    $$sum_{i=k}^infty(-1)^{i-k}egin{bmatrix}i \ kend{bmatrix}frac{x^i}{i!}=frac{ln(x+1)^k}{k!}$$

    这里先贴一个90分代码,之后再卡常。。。

      1 #include<bits/stdc++.h>
      2 #define Rint register int
      3 using namespace std;
      4 typedef long long LL;
      5 const int mod = 167772161, G = 3, Gi = 55924054, N = 1 << 19;
      6 inline int kasumi(int a, int b){
      7     int res = 1;
      8     while(b){
      9         if(b & 1) res = (LL) res * a % mod;
     10         a = (LL) a * a % mod;
     11         b >>= 1;
     12     }
     13     return res;
     14 }
     15 int fac[N], inv[N], invfac[N];
     16 inline void init(int n){
     17     fac[0] = 1;
     18     for(Rint i = 1;i <= n;i ++) fac[i] = (LL) i * fac[i - 1] % mod;
     19     invfac[n] = kasumi(fac[n], mod - 2);
     20     for(Rint i = n;i;i --){
     21         invfac[i - 1] = (LL) i * invfac[i] % mod;
     22         inv[i] = (LL) invfac[i] * fac[i - 1] % mod;
     23     }
     24 }
     25 int rev[N];
     26 inline int calrev(int n){
     27     int limit = 1, L = -1;
     28     while(limit <= n){limit <<= 1; L ++;}
     29     for(Rint i = 0;i < limit;i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
     30     return limit;
     31 }
     32 inline void NTT(int *A, int limit, int type){
     33     for(Rint i = 0;i < limit;i ++) if(i < rev[i]) swap(A[i], A[rev[i]]);
     34     for(Rint mid = 1;mid < limit;mid <<= 1){
     35         int Wn = kasumi(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
     36         for(Rint j = 0;j < limit;j += (mid << 1)){
     37             int w = 1;
     38             for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
     39                 int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
     40                 A[j + k] = (x + y) % mod;
     41                 A[j + k + mid] = (x - y + mod) % mod;
     42             }
     43         }
     44     }
     45     if(type == -1){
     46         int inv = kasumi(limit, mod - 2);
     47         for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
     48     }
     49 }
     50 int ans[N];
     51 inline void poly_inv(int *A, int deg){
     52     static int tmp[N];
     53     if(deg == 1){
     54         ans[0] = kasumi(A[0], mod - 2);
     55         return;
     56     }
     57     poly_inv(A, deg + 1 >> 1);
     58     int limit = calrev(deg << 1);
     59     for(Rint i = 0;i < deg;i ++) tmp[i] = A[i];
     60     for(Rint i = deg;i < limit;i ++) tmp[i] = ans[i] = 0;
     61     NTT(tmp, limit, 1); NTT(ans, limit, 1);
     62     for(Rint i = 0;i < limit;i ++) ans[i] = (2 - (LL) ans[i] * tmp[i] % mod + mod) * ans[i] % mod;
     63     NTT(ans, limit, -1);
     64     for(Rint i = deg;i < limit;i ++) ans[i] = 0;
     65 }
     66 int Ln[N];
     67 inline void poly_Ln(int *A, int deg){
     68     static int tmp[N];
     69     if(deg == 1){
     70         Ln[0] = 0;
     71         return;
     72     }
     73     poly_inv(A, deg);
     74     int limit = calrev(deg << 1);
     75     for(Rint i = 1;i < deg;i ++) tmp[i - 1] = (LL) i * A[i] % mod;
     76     for(Rint i = deg - 1;i < limit;i ++) tmp[i] = 0;
     77     NTT(ans, limit, 1); NTT(tmp, limit, 1);
     78     for(Rint i = 0;i < limit;i ++) Ln[i] = (LL) ans[i] * tmp[i] % mod;
     79     NTT(Ln, limit, -1);
     80     for(Rint i = deg + 1;i < limit;i ++) Ln[i] = 0;
     81     for(Rint i = deg;i;i --) Ln[i] = (LL) Ln[i - 1] * inv[i] % mod;
     82     Ln[0] = 0;
     83     for(Rint i = 0;i < limit;i ++) ans[i] = tmp[i] = 0;
     84 }
     85 int Exp[N];
     86 inline void poly_Exp(int *A, int deg){
     87     if(deg == 1){
     88         Exp[0] = 1;
     89         return;
     90     }
     91     poly_Exp(A, deg + 1 >> 1);
     92     poly_Ln(Exp, deg);
     93     for(Rint i = 0;i < deg;i ++) Ln[i] = (A[i] + (i == 0) - Ln[i] + mod) % mod;
     94     int limit = calrev(deg << 1);
     95     NTT(Exp, limit, 1); NTT(Ln, limit, 1);
     96     for(Rint i = 0;i < limit;i ++) Exp[i] = (LL) Exp[i] * Ln[i] % mod;
     97     NTT(Exp, limit, -1);
     98     for(Rint i = deg;i < limit;i ++) Exp[i] = ans[i] = Ln[i] = 0;
     99     for(Rint i = 0;i < deg;i ++) ans[i] = Ln[i] = 0;
    100 }
    101 int n, k, A[N];
    102 int main(){
    103     scanf("%d%d", &n, &k);
    104     if(n < k){
    105         for(Rint i = 0;i <= n;i ++) printf("0 ");
    106         return 0;
    107     }
    108     init(n);
    109     A[0] = A[1] = 1;
    110     poly_Ln(A, n + 1);
    111 //    for(Rint i = 0;i <= n;i ++) printf("%d ", Ln[i]);
    112 //    putchar('
    ');
    113     int lst = -1, val = 0;
    114     for(Rint i = 0;i <= n;i ++) if(Ln[i]) {val = Ln[lst = i]; break;}
    115     for(Rint i = 0;i <= n - lst;i ++) A[i] = (LL) Ln[i + lst] * inv[val] % mod;
    116     poly_Ln(A, n + 1);
    117     for(Rint i = 0;i <= n;i ++) A[i] = (LL) k * Ln[i] % mod;
    118     poly_Exp(A, n + 1);
    119     val = kasumi(val, k);
    120     for(Rint i = 0;i < lst * k;i ++) A[i] = 0;
    121     for(Rint i = 0;i <= n - lst * k;i ++) A[i + lst * k] = (LL) val * Exp[i] % mod * invfac[k] % mod;
    122     for(Rint i = 0;i <= n;i ++)
    123         printf("%d ", (LL) A[i] * fac[i] % mod * ((i - k) & 1 ? (mod - 1) : 1) % mod);
    124 }
    Luogu5409

    第一类斯特林数-行

    $$sum_{i=0}^{+infty}egin{bmatrix}n \ iend{bmatrix}x^i=prod_{i=0}^{n-1}(x+i)$$

    我们定义$f_n(x)=prod_{i=0}^{n-1}(x+i)$.

    所以$f_{2n}(x)=f_n(x)f_n(x+n)$,我们考虑多项式与一次函数的复合。

    若$f_n(x)=sum_{i=0}^na_ix^i$则

    $$f_n(x+n)=sum_{i=0}^na_i(x+c)^i$$

    $$=sum_{i=0}^na_isum_{j=0}^ix^jc^{i-j}inom{i}{j}$$

    $$=sum_{j=0}^nfrac{x^j}{j!}sum_{i=j}^n(a_i*i!)*frac{c^{i-j}}{(i-j)!}$$

    直接卷积就可以。

    (我连对着式子看都可以写错,我服了自己了。。。)

     1 #include<bits/stdc++.h>
     2 #define Rint register int
     3 using namespace std;
     4 typedef long long LL;
     5 const int N = 1 << 19, mod = 167772161, g = 3, gi = 55924054;
     6 inline int kasumi(int a, int b){
     7     int res = 1;
     8     while(b){
     9         if(b & 1) res = (LL) res * a % mod;
    10         a = (LL) a * a % mod;
    11         b >>= 1;
    12     }
    13     return res;
    14 }
    15 int fac[N], inv[N], invfac[N];
    16 inline void init(int n){
    17     fac[0] = 1;
    18     for(Rint i = 1;i <= n;i ++) fac[i] = (LL) i * fac[i - 1] % mod;
    19     invfac[n] = kasumi(fac[n], mod - 2);
    20     for(Rint i = n;i;i --){
    21         invfac[i - 1] = (LL) i * invfac[i] % mod;
    22         inv[i] = (LL) fac[i - 1] * invfac[i] % mod;
    23     }
    24 }
    25 int rev[N];
    26 inline int calrev(int n){
    27     int limit = 1, L = -1;
    28     while(limit <= n){limit <<= 1; L ++;}
    29     for(Rint i = 0;i < limit;i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    30     return limit;
    31 }
    32 inline void NTT(int *A, int limit, int type){
    33     for(Rint i = 0;i < limit;i ++) if(i < rev[i]) swap(A[i], A[rev[i]]);
    34     for(Rint mid = 1;mid < limit;mid <<= 1){
    35         int Wn = kasumi(type == 1 ? g : gi, (mod - 1) / (mid << 1));
    36         for(Rint j = 0;j < limit;j += (mid << 1)){
    37             int w = 1;
    38             for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
    39                 int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
    40                 A[j + k] = (x + y) % mod;
    41                 A[j + k + mid] = (x - y + mod) % mod;
    42             }
    43         }
    44     }
    45     if(type == -1){
    46         int inv = kasumi(limit, mod - 2);
    47         for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * inv % mod;
    48     }
    49 }
    50 int F[N], G[N], tmp[N];
    51 inline void work(int n){
    52     if(n == 1){
    53         F[0] = 0; F[1] = 1;
    54         return;
    55     }
    56     int mid = n >> 1; work(mid);
    57     for(Rint i = 0, now = 1;i <= mid;i ++, now = (LL) now * mid % mod){
    58         tmp[i] = (LL) F[i] * fac[i] % mod;
    59         G[mid - i] = (LL) now * invfac[i] % mod;
    60     }
    61     int limit = calrev(n);
    62     NTT(tmp, limit, 1); NTT(G, limit, 1);
    63     for(Rint i = 0;i < limit;i ++) tmp[i] = (LL) G[i] * tmp[i] % mod;
    64     NTT(tmp, limit, -1);
    65     for(Rint i = 0;i <= mid;i ++)
    66         G[i] = (LL) tmp[mid + i] * invfac[i] % mod;
    67     for(Rint i = mid + 1;i < limit;i ++) F[i] = G[i] = 0;
    68     NTT(F, limit, 1); NTT(G, limit, 1);
    69     for(Rint i = 0;i < limit;i ++) F[i] = (LL) F[i] * G[i] % mod;
    70     NTT(F, limit, -1);
    71     for(Rint i = (mid << 1) + 1;i < limit;i ++) F[i] = 0;
    72     for(Rint i = 0;i < limit;i ++) G[i] = tmp[i] = 0;
    73     if(n & 1){
    74         for(Rint i = 0;i <= n;i ++) tmp[i] = (LL) (n - 1) * F[i] % mod;
    75         for(Rint i = 0;i < n;i ++) tmp[i + 1] = (tmp[i + 1] + F[i]) % mod;
    76         for(Rint i = 0;i <= n;i ++) F[i] = tmp[i];
    77     }
    78 }
    79 int n;
    80 int main(){
    81     scanf("%d", &n);
    82     init(n); work(n);
    83     for(Rint i = 0;i <= n;i ++) printf("%d ", F[i]);
    84 }
    Luogu5408
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  • 原文地址:https://www.cnblogs.com/AThousandMoons/p/10943784.html
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