bzoj 4555 求和

题目大意:

求$\sum\limits_{i=0}^n \sum\limits_{j=0}^i S2(i,j) \times 2^j \times j!$

思路:

法1:

首先把斯特林数展开$S2(i,j)=\frac{1}{j!} \sum\limits_{k=0}^j (-1)^k * \binom{j}{k} * (j-k)^i$

然后先交换求和顺序提出只与$j$有关的项,同时把$j$的范围可以扩展到$n$

代入得到:$\sum\limits_{j=0}^n 2^j * j! \sum\limits_{i=0}^n \sum\limits_{k=0}^j \frac{(-1)^k}{j!} * \binom{j}{k} * (j-k)^i$ 

拆开组合数:$\sum\limits_{j=0}^n 2^j * j! \sum\limits_{i=0}^n \sum\limits_{k=0}^j \frac{(-1)^k}{k!} * \frac{(j-k)^i}{(j-k)!}$ 

把只于$k$有关的提出来:$\sum\limits_{j=0}^n 2^j * j! \sum\limits_{k=0}^j \frac{(-1)^k}{k!} * \frac{\sum\limits_{i=0}^n (j-k)^i}{(j-k)!}$ 

设$f(i)=\frac {(-1)^i}{i!},g(i)=\frac{\sum\limits_{j=0}^n i^j}{i!}$

则原式为$\sum\limits_{i=0}^n 2^i \times (i!) \times (f*g)(i)$

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 200100
15 #define MOD 998244353
16 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
17 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
18 #define ren for(register int i=fst[x];i;i=nxt[i])
19 #define pb(i,x) vec[i].push_back(x)
20 #define pls(a,b) ((a+b)%MOD+MOD)%MOD
21 #define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
22 #define mul(a,b) (1LL*(a)*(b))%MOD
23 using namespace std;
24 inline int read()
25 {
26     int x=0,f=1;char ch=getchar();
27     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
28     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
29     return x*f;
30 }
31 int n,a[MAXN<<1],b[MAXN<<1],fac[MAXN],ifac[MAXN];
32 int rev[MAXN<<1],lim,lg,pw[30],ipw[30],ans;
33 int q_pow(int bas,int t,int res=1)
34 {
35     for(;t;t>>=1,bas=mul(bas,bas)) if(t&1) res=mul(res,bas);return res;
36 }
37 #define inv(x) q_pow(x,MOD-2)
38 void ntt(int *a,int n,int f)
39 {
40     rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
41     for(int i=1,t=1;i<n;i<<=1,++t)
42     {
43         int wn= f>0?pw[t]:ipw[t];for(int j=0;j<n;j+=i<<1)
44         {
45             int w=1,x,y;for(int k=0;k<i;++k,w=mul(w,wn))
46                 x=a[j+k],y=mul(a[j+k+i],w),a[j+k]=pls(x,y),a[j+k+i]=mns(x,y);
47         }
48     }
49     if(f>0) return ;int nv=inv(n);rep(i,0,n-1) a[i]=mul(a[i],nv);
50 }
51 int main()
52 {
53     n=read();fac[0]=ifac[0]=1;rep(i,1,n) fac[i]=mul(fac[i-1],i),ifac[i]=inv(fac[i]);
54     int tmp=-1;rep(i,0,n) tmp=(MOD-tmp)%MOD,a[i]=mul(tmp,ifac[i]);
55     b[0]=1,b[1]=n+1;rep(i,2,n) b[i]=mul(ifac[i],mul(inv(i-1),(q_pow(i,n+1)-1)));
56     for(lim=1,lg=1;lim<=(n+1<<1);lim<<=1,lg++)
57         pw[lg]=q_pow(3,(MOD-1)/(1<<lg)),ipw[lg]=inv(pw[lg]);
58     rep(i,1,lim-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<lg-2);
59     ntt(a,lim,1);ntt(b,lim,1);rep(i,0,lim-1) a[i]=mul(a[i],b[i]);ntt(a,lim,-1);
60     tmp=1;rep(i,0,n) ans=pls(ans,mul(mul(a[i],tmp),fac[i])),tmp=pls(tmp,tmp);
61     printf("%d\n",ans);
62 }
View Code

 

法2:

思考式子的组合意义,考虑$f(i)=\sum\limits_{j=0}^i S2(i,j) \times 2^j \times j!$

即表示将$i$个数分成$j$个有序集合且每个集合有两种状态的所有方案

那么我们可以枚举最后一个集合中数的数量即$f(n)=\sum\limits_{i=1}^n 2 * \binom{n}{i}* f(n-i)$

两边同除以阶乘,然后就可以分治或者求逆了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<vector>
 9 #include<map>
10 #include<set>
11 #define ll long long
12 #define db double
13 #define inf 2139062143
14 #define MAXN 200100
15 #define MOD 998244353
16 #define rep(i,s,t) for(register int i=(s),i##__end=(t);i<=i##__end;++i)
17 #define dwn(i,s,t) for(register int i=(s),i##__end=(t);i>=i##__end;--i)
18 #define ren for(register int i=fst[x];i;i=nxt[i])
19 #define pb(i,x) vec[i].push_back(x)
20 #define pls(a,b) ((a+b)%MOD+MOD)%MOD
21 #define mns(a,b) ((a%MOD-(b)%MOD)%MOD+MOD)%MOD
22 #define mul(a,b) (1LL*(a)*(b))%MOD
23 using namespace std;
24 inline int read()
25 {
26     int x=0,f=1;char ch=getchar();
27     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
28     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
29     return x*f;
30 }
31 int n,a[MAXN<<1],b[MAXN<<1],fac[MAXN],ifac[MAXN],iv[MAXN<<1],tmp[MAXN<<1];
32 int rev[MAXN<<1],lim,lg,pw[30],ipw[30],ans,l2[MAXN<<2];
33 int q_pow(int bas,int t,int res=1)
34 {
35     for(;t;t>>=1,bas=mul(bas,bas)) if(t&1) res=mul(res,bas);return res;
36 }
37 #define inv(x) q_pow(x,MOD-2)
38 void ntt(int *a,int n,int f)
39 {
40     rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
41     for(int i=1,t=1;i<n;i<<=1,++t)
42     {
43         int wn= f>0?pw[t]:ipw[t];for(int j=0;j<n;j+=i<<1)
44         {
45             int w=1,x,y;for(int k=0;k<i;++k,w=mul(w,wn))
46                 x=a[j+k],y=mul(a[j+k+i],w),a[j+k]=pls(x,y),a[j+k+i]=mns(x,y);
47         }
48     }
49     if(f>0) return ;int nv=inv(n);rep(i,0,n-1) a[i]=mul(a[i],nv);
50 }
51 void solve(int *a,int *b,int lmt)
52 {
53     rep(i,1,lmt-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<l2[lmt]-1);
54     ntt(a,lmt,1);ntt(b,lmt,1);
55     rep(i,0,lmt-1) a[i]=mul(mns(2,mul(a[i],b[i])),a[i]);ntt(a,lmt,-1);
56 }
57 void Inv(int *a,int *f,int lmt)
58 {
59     f[0]=inv(a[0]);for(int t=2;t<=lmt;t<<=1)
60         {rep(i,0,t-1)tmp[i]=a[i];solve(f,tmp,t<<1);rep(i,t,(t<<1)-1) f[i]=0;}
61 }
62 int main()
63 {
64     n=read();fac[0]=ifac[0]=1;rep(i,1,n) fac[i]=mul(fac[i-1],i),ifac[i]=inv(fac[i]);
65     for(lim=1,lg=1;lim<=(n+1)<<1;lim<<=1,lg++)
66         l2[1<<lg]=lg,pw[lg]=q_pow(3,(MOD-1)/(1<<lg)),ipw[lg]=inv(pw[lg]);
67     lim>>=1;rep(i,1,n) b[i]=mns(MOD,mul(2,ifac[i]));b[0]=pls(b[0],1);
68     Inv(b,a,lim);rep(i,0,n) ans=pls(ans,mul(a[i],fac[i]));printf("%d\n",ans);
69 }
View Code

 

posted @ 2019-04-03 16:51  jack_yyc  阅读(82)  评论(0编辑  收藏  举报