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

设

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

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

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

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