luogu4389 付公主的背包

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 }

luogu4389

 

转载于:https://www.cnblogs.com/AThousandMoons/p/10524935.html