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链式反应
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鏈式反應
題目描述
不想看題面，其實就是給定p(x)p(x)p(x)，有f′(x)=f(x)2p(x)+1f'(x)=f(x)^2p(x)+1f′(x)=f(x)2p(x)+1，求fff這個多項式的冪級數形式前nnn項。
Solution
式子可以寫成fn=∑i,j[0<=i+j<n]fifjpn?i?j?1f_n=\sum_{i,j}[0<=i+j<n]f_if_jp_{n-i-j-1}fn?=∑i,j?[0<=i+j<n]fi?fj?pn?i?j?1?。
顯然能分治FFTFFTFFT。
設分治區(qū)間為[l,r][l,r][l,r]，考慮[l,mid][l,mid][l,mid]對[mid+1,r][mid+1,r][mid+1,r]的貢獻。
當l=1l=1l=1時，貢獻為
∑i=0mid∑j=0mid∑k=0rfifjpk\sum^{mid}_{i=0}\sum^{mid}_{j=0}\sum^{r}_{k=0}f_if_jp_k i=0∑mid?j=0∑mid?k=0∑r?fi?fj?pk?
當l>1l>1l>1時，有2l>r2l>r2l>r，因此貢獻為
2∑i=lmid∑j=0r?l∑k=0r?lfifjpk2\sum^{mid}_{i=l}\sum^{r-l}_{j=0}\sum^{r-l}_{k=0}f_if_jp_k 2i=l∑mid?j=0∑r?l?k=0∑r?l?fi?fj?pk?
時間復雜度兩只lglglg。
注意這里的f,pf,pf,p始終是一個EGFEGFEGF。
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std
;template<typename T
>inline bool upmin(T
&x
,T y
) { return y
<x
?x
=y
,1:0; }
template<typename T
>inline bool upmax(T
&x
,T y
) { return x
<y
?x
=y
,1:0; }typedef long long ll
;
typedef unsigned long long ull
;
typedef long double lod
;
typedef pair
<int,int> PR
;
typedef vector
<int> VI
;const lod eps
=1e-11;
const lod pi
=acos(-1);
const int oo
=1<<30;
const ll loo
=1ll<<62;
const int mods
=998244353;
const int g
=3;
const int gi
=(mods
+1)/3;
const int MAXN
=2000005;
const int INF
=0x3f3f3f3f;
inline int read()
{int f
=1,x
=0; char c
=getchar();while (c
<'0'||c
>'9') { if (c
=='-') f
=-1; c
=getchar(); }while (c
>='0'&&c
<='9') { x
=(x
<<3)+(x
<<1)+(c
^48); c
=getchar(); }return x
*f
;
}
int Limit
,L
;
char st
[MAXN
];
int F
[MAXN
],G
[MAXN
],P
[MAXN
],H
[MAXN
],Q
[MAXN
],rev
[MAXN
],inv
[MAXN
],fac
[MAXN
],n
;
inline int upd(int x
,int y
) { return (x
+y
>=mods
)?x
+y
-mods
:x
+y
; }
inline int quick_pow(int x
,int y
)
{int ret
=1;for (;y
;y
>>=1){if (y
&1) ret
=1ll*ret
*x
%mods
;x
=1ll*x
*x
%mods
;}return ret
;
}
inline void Init(int n
)
{fac
[0]=1;for (int i
=1;i
<=n
;i
++) fac
[i
]=1ll*fac
[i
-1]*i
%mods
;inv
[n
]=quick_pow(fac
[n
],mods
-2);for (int i
=n
-1;i
>=0;i
--) inv
[i
]=1ll*inv
[i
+1]*(i
+1)%mods
;
}
void Number_Theoretic_Transform(int *A
,int opt
)
{for (int i
=0;i
<Limit
;i
++) if (i
<rev
[i
]) swap(A
[i
],A
[rev
[i
]]);for (int i
=1;i
<Limit
;i
<<=1){int Wn
=quick_pow(opt
==1?g
:gi
,(mods
-1)/(i
<<1));for (int j
=0;j
<Limit
;j
+=(i
<<1))for (int k
=j
,w
=1;k
<j
+i
;k
++,w
=1ll*w
*Wn
%mods
){int x
=A
[k
],y
=1ll*A
[k
+i
]*w
%mods
;A
[k
]=upd(x
,y
),A
[k
+i
]=upd(x
,mods
-y
);}}if (opt
==-1){int invlim
=quick_pow(Limit
,mods
-2);for (int i
=0;i
<Limit
;i
++) A
[i
]=1ll*A
[i
]*invlim
%mods
;}
}
void solve(int l
,int r
)
{if (l
==r
) { if (l
==1) F
[l
]=(mods
+1)>>1;else F
[l
]=1ll*F
[l
]*inv
[l
]%mods
*fac
[l
-1]%mods
;printf("%d\n",2ll*F
[l
]*fac
[l
]%mods
);return; }int mid
=(l
+r
)>>1;solve(l
,mid
);if (l
==1){Limit
=1,L
=0;int len
=mid
*2+r
;while (Limit
<=len
) Limit
<<=1,L
++;for (int i
=0;i
<Limit
;i
++) rev
[i
]=(rev
[i
>>1]>>1)|((i
&1)<<(L
-1));for (int i
=0;i
<Limit
;i
++) H
[i
]=G
[i
]=0;for (int i
=0;i
<=r
;i
++) H
[i
]=P
[i
];for (int i
=0;i
<=mid
;i
++) G
[i
]=F
[i
];Number_Theoretic_Transform(G
,1);Number_Theoretic_Transform(H
,1);for (int i
=0;i
<Limit
;i
++) G
[i
]=1ll*G
[i
]*G
[i
]%mods
*H
[i
]%mods
;Number_Theoretic_Transform(G
,-1);for (int i
=mid
+1;i
<=r
;i
++) F
[i
]=upd(F
[i
],G
[i
]);}else{Limit
=1,L
=0;int len
=(mid
-l
)+(r
-l
)*2;while (Limit
<=len
) Limit
<<=1,L
++;for (int i
=0;i
<Limit
;i
++) rev
[i
]=(rev
[i
>>1]>>1)|((i
&1)<<(L
-1));for (int i
=0;i
<Limit
;i
++) H
[i
]=G
[i
]=Q
[i
]=0;for (int i
=0;i
<=r
-l
;i
++) H
[i
]=P
[i
];for (int i
=0;i
<=mid
-l
;i
++) G
[i
]=F
[i
+l
];for (int i
=0;i
<=r
-l
;i
++) Q
[i
]=F
[i
];Number_Theoretic_Transform(G
,1);Number_Theoretic_Transform(Q
,1);Number_Theoretic_Transform(H
,1);for (int i
=0;i
<Limit
;i
++) G
[i
]=1ll*G
[i
]*Q
[i
]%mods
*H
[i
]%mods
;Number_Theoretic_Transform(G
,-1);for (int i
=mid
+1;i
<=r
;i
++) F
[i
]=upd(F
[i
],upd(G
[i
-l
],G
[i
-l
]));}solve(mid
+1,r
);
}
int main()
{n
=read();Init(n
);scanf("%s",st
);for (int i
=1;i
<=n
;i
++) P
[i
]=(st
[i
-1]-'0')*inv
[i
-1];solve(1,n
);return 0;
}
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