题目链接：洛谷
题目大意：现在有$n$个物品，每种物品体积为$v_i$，对任意$s\in [1,m]$，求背包恰好装$s$体积的方案数（完全背包问题）。
数据范围：$n,m\leq 10^5$
这道题，看到数据范围就知道是生成函数。
$$Ans=\prod_{i=1}^n\frac{1}{1-x^{v_i}}$$
但是这个式子直接乘会tle，我们考虑进行优化。
看见这个连乘的式子，应该是要上$\ln$.
$$Ans=\exp(\sum_{i=1}^n\ln(\frac{1}{1-x^{v_i}}))$$
接下来的问题就是如何快速计算$\ln(\frac{1}{1-x^{v_i}})$。
$$\ln(f(x))=\int f'f^{-1}dx$$
所以
$$\ln(\frac{1}{1-x^v})=\int\sum_{i=1}^{+\infty}vix^{vi-1}*(1-x^v)dx$$
$$=\int(\sum_{i=1}^{+\infty}vix^{vi-1}-\sum_{i=2}^{+\infty}v(i-1)x^{vi-1})dx$$
$$=\int(\sum_{i=1}^{+\infty}vx^{vi-1})dx$$
$$=\sum_{i=1}^{+\infty}\frac{1}{i}x^{vi}$$
然后就可以直接代公式了。
1 #include<cstdio>
2 #include<algorithm>
3 #define Rint register int
4 using namespace std;
5 typedef long long LL;
6 const int N = 400003, P = 998244353, G = 3, Gi = 332748118;
7 int n, m, cnt[N], A[N];
8 inline int kasumi(int a, int b){
9 int res = 1;
10 while(b){
11 if(b & 1) res = (LL) res * a % P;
12 a = (LL) a * a % P;
13 b >>= 1;
14 }
15 return res;
16 }
17 int R[N];
18 inline void NTT(int *A, int limit, int type){
19 for(Rint i = 1;i < limit;i ++)
20 if(i < R[i]) swap(A[i], A[R[i]]);
21 for(Rint mid = 1;mid < limit;mid <<= 1){
22 int Wn = kasumi(type == 1 ? G : Gi, (P - 1) / (mid << 1));
23 for(Rint j = 0;j < limit;j += mid << 1){
24 int w = 1;
25 for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % P){
26 int x = A[j + k], y = (LL) w * A[j + k + mid] % P;
27 A[j + k] = (x + y) % P;
28 A[j + k + mid] = (x - y + P) % P;
29 }
30 }
31 }
32 if(type == -1){
33 int inv = kasumi(limit, P - 2);
34 for(Rint i = 0;i < limit;i ++)
35 A[i] = (LL) A[i] * inv % P;
36 }
37 }
38 int ans[N];
39 inline void poly_inv(int *A, int deg){
40 static int tmp[N];
41 if(deg == 1){
42 ans[0] = kasumi(A[0], P - 2);
43 return;
44 }
45 poly_inv(A, (deg + 1) >> 1);
46 int limit = 1, L = -1;
47 while(limit <= (deg << 1)){limit <<= 1; L ++;}
48 for(Rint i = 1;i < limit;i ++)
49 R[i] = (R[i >> 1] >> 1) | ((i & 1) << L);
50 for(Rint i = 0;i < deg;i ++) tmp[i] = A[i];
51 for(Rint i = deg;i < limit;i ++) tmp[i] = 0;
52 NTT(tmp, limit, 1); NTT(ans, limit, 1);
53 for(Rint i = 0;i < limit;i ++)
54 ans[i] = (2 - (LL) tmp[i] * ans[i] % P + P) % P * ans[i] % P;
55 NTT(ans, limit, -1);
56 for(Rint i = deg;i < limit;i ++) ans[i] = 0;
57 }
58 int Ln[N];
59 inline void get_Ln(int *A, int deg){
60 static int tmp[N];
61 poly_inv(A, deg);
62 for(Rint i = 1;i < deg;i ++)
63 tmp[i - 1] = (LL) i * A[i] % P;
64 tmp[deg - 1] = 0;
65 int limit = 1, L = -1;
66 while(limit <= (deg << 1)){limit <<= 1; L ++;}
67 for(Rint i = 1;i < limit;i ++)
68 R[i] = (R[i >> 1] >> 1) | ((i & 1) << L);
69 NTT(ans, limit, 1); NTT(tmp, limit, 1);
70 for(Rint i = 0;i < limit;i ++) Ln[i] = (LL) ans[i] * tmp[i] % P;
71 NTT(Ln, limit, -1);
72 for(Rint i = deg + 1;i < limit;i ++) Ln[i] = 0;
73 for(Rint i = deg;i;i --) Ln[i] = (LL) Ln[i - 1] * kasumi(i, P - 2) % P;
74 for(Rint i = 0;i < limit;i ++) tmp[i] = ans[i] = 0;
75 Ln[0] = 0;
76 }
77 int Exp[N];
78 inline void get_Exp(int *A, int deg){
79 if(deg == 1){
80 Exp[0] = 1;
81 return;
82 }
83 get_Exp(A, (deg + 1) >> 1);
84 get_Ln(Exp, deg);
85 for(Rint i = 0;i < deg;i ++) Ln[i] = (A[i] + (i == 0) - Ln[i] + P) % P;
86 int limit = 1, L = -1;
87 while(limit <= (deg << 1)){limit <<= 1; L ++;}
88 for(Rint i = 1;i < limit;i ++)
89 R[i] = (R[i >> 1] >> 1) | ((i & 1) << L);
90 NTT(Exp, limit, 1); NTT(Ln, limit, 1);
91 for(Rint i = 0;i < limit;i ++) Exp[i] = (LL) Exp[i] * Ln[i] % P;
92 NTT(Exp, limit, -1);
93 for(Rint i = deg;i < limit;i ++) Exp[i] = 0;
94 for(Rint i = 0;i < limit;i ++) Ln[i] = ans[i] = 0;
95 }
96 int main(){
97 scanf("%d%d", &n, &m);
98 for(Rint i = 1;i <= n;i ++){
99 int x;
100 scanf("%d", &x);
101 ++ cnt[x];
102 }
103 for(Rint i = 1;i <= m;i ++){
104 if(!cnt[i]) continue;
105 for(Rint j = i;j <= m;j += i)
106 A[j] = (A[j] + (LL) cnt[i] * kasumi(j / i, P - 2) % P) % P;
107 }
108 get_Exp(A, m + 1);
109 for(Rint i = 1;i <= m;i ++)
110 printf("%d\n", Exp[i]);
111 }
来源：https://www.cnblogs.com/AThousandMoons/p/10524935.html