題目鏈接：洛谷

題目大意：現在有$n$個物品，每種物品體積為$v_i$，對任意$s\in [1,m]$，求背包恰好裝$s$體積的方案數（完全背包問題）。

數據范圍：$n,m\leq 10^5$

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這道題，看到數據范圍就知道是生成函數。
$$Ans=\prod_{i=1}^n\frac{1}{1-x^{v_i}}$$

但是這個式子直接乘會tle，我們考慮進行優化。

看見這個連乘的式子，應該是要上$\ln$.

$$Ans=\exp(\sum_{i=1}^n\ln(\frac{1}{1-x^{v_i}}))$$

接下來的問題就是如何快速計算$\ln(\frac{1}{1-x^{v_i}})$。

$$\ln(f(x))=\int f'f^{-1}dx$$

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 } luogu4389

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轉載于:https://www.cnblogs.com/AThousandMoons/p/10524935.html

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