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HDU 5730 Shell Necklace（生成函数 多项式求逆）
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Shell Necklace
由題意可得f[n]=∑i=1na[i]f[n?i]f[n] = \sum\limits_{i = 1} ^{n} a[i] f[n - i]f[n]=i=1∑n?a[i]f[n?i]，設f[n]f[n]f[n]的生成函數為F(x)F(x)F(x)，a[n]a[n]a[n]的生成函數為A(n)A(n)A(n)
F(x)A(x)=∑n≥0xn∑i+j=naifn?i由于a0=0，a0f0=0，右側卷積缺了f0這一項F(x)A(x)=F(x)?f0F(x)=11?A(x)多項式求個逆就好了F(x)A(x) = \sum_{n \geq 0} x ^ n \sum_{i + j = n} a_if_{n - i}\\ 由于a_0 = 0，a_0 f_0 = 0，右側卷積缺了f_0這一項\\ F(x) A(x) = F(x) - f_0\\ F(x) = \frac{1}{1 - A(x)}\\ 多項式求個逆就好了\\ F(x)A(x)=n≥0∑?xni+j=n∑?ai?fn?i?由于a0?=0，a0?f0?=0，右側卷積缺了f0?這一項F(x)A(x)=F(x)?f0?F(x)=1?A(x)1?多項式求個逆就好了
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std
;
#define ll long long
#define ull unsigned long long
#define RG register
#define MAX 288888
#define MOD (313)
const double Pi
=acos(-1);
const int m
=sqrt(MOD
);
inline int read()
{RG
int x
=0,t
=1;RG
char ch
=getchar();while((ch
<'0'||ch
>'9')&&ch
!='-')ch
=getchar();if(ch
=='-')t
=-1,ch
=getchar();while(ch
<='9'&&ch
>='0')x
=x
*10+ch
-48,ch
=getchar();return x
*t
;
}
int fpow(int a
,int b
){int s
=1;while(b
){if(b
&1)s
=1ll*s
*a
%MOD
;a
=1ll*a
*a
%MOD
;b
>>=1;}return s
;}
struct Complex
{double a
,b
;}W
[MAX
],A
[MAX
],B
[MAX
],C
[MAX
],D
[MAX
];
Complex
operator+(Complex a
,Complex b
){return (Complex
){a
.a
+b
.a
,a
.b
+b
.b
};}
Complex
operator-(Complex a
,Complex b
){return (Complex
){a
.a
-b
.a
,a
.b
-b
.b
};}
Complex
operator*(Complex a
,Complex b
){return (Complex
){a
.a
*b
.a
-a
.b
*b
.b
,a
.a
*b
.b
+a
.b
*b
.a
};}
int r
[MAX
];
void FFT(Complex
*P
,int N
,int opt
)
{for(int i
=0;i
<N
;++i
)if(i
<r
[i
])swap(P
[i
],P
[r
[i
]]);for(int i
=1;i
<N
;i
<<=1)for(int p
=i
<<1,j
=0;j
<N
;j
+=p
)for(int k
=0;k
<i
;++k
){Complex w
=(Complex
){W
[N
/i
*k
].a
,W
[N
/i
*k
].b
*opt
};Complex X
=P
[j
+k
],Y
=P
[i
+j
+k
]*w
;P
[j
+k
]=X
+Y
;P
[i
+j
+k
]=X
-Y
;}if(opt
==-1)for(int i
=0;i
<N
;++i
)P
[i
].a
/=1.0*N
;
}
void Multi(int *a
,int *b
,int len
,int *ret
)
{for(int i
=0;i
<(len
<<1);++i
)A
[i
]=B
[i
]=C
[i
]=D
[i
]=(Complex
){0,0};for(int i
=0;i
<len
;++i
){a
[i
]%=MOD
;b
[i
]%=MOD
;A
[i
]=(Complex
){(a
[i
]/m
)*1.0,0};B
[i
]=(Complex
){(a
[i
]%m
)*1.0,0};C
[i
]=(Complex
){(b
[i
]/m
)*1.0,0};D
[i
]=(Complex
){(b
[i
]%m
)*1.0,0};}int N
,l
=0;for(N
=1;N
<=len
;N
<<=1)++l
;for(int i
=0;i
<N
;++i
)r
[i
]=(r
[i
>>1]>>1)|((i
&1)<<(l
-1));for(int i
=1;i
<N
;i
<<=1)for(int k
=0;k
<i
;++k
)W
[N
/i
*k
]=(Complex
){cos(k
*Pi
/i
),sin(k
*Pi
/i
)};FFT(A
,N
,1);FFT(B
,N
,1);FFT(C
,N
,1);FFT(D
,N
,1);for(int i
=0;i
<N
;++i
){Complex tmp
=A
[i
]*C
[i
];C
[i
]=B
[i
]*C
[i
],B
[i
]=B
[i
]*D
[i
],D
[i
]=D
[i
]*A
[i
];A
[i
]=tmp
;C
[i
]=C
[i
]+D
[i
];}FFT(A
,N
,-1);FFT(B
,N
,-1);FFT(C
,N
,-1);for(int i
=0;i
<len
;++i
){ret
[i
]=0;ret
[i
]=(ret
[i
]+1ll*(ll
)(A
[i
].a
+0.5)%MOD
*m
%MOD
*m
%MOD
)%MOD
;ret
[i
]=(ret
[i
]+1ll*(ll
)(C
[i
].a
+0.5)%MOD
*m
%MOD
)%MOD
;ret
[i
]=(ret
[i
]+1ll*(ll
)(B
[i
].a
+0.5)%MOD
)%MOD
;ret
[i
]=(ret
[i
]+MOD
)%MOD
;}
}
int c
[MAX
],d
[MAX
];
void Inv(int *a
,int *b
,int len
)
{if(len
==1){b
[0]=fpow(a
[0],MOD
-2);return;}Inv(a
,b
,len
>>1);Multi(a
,b
,len
,c
);Multi(c
,b
,len
,d
);for(int i
=0;i
<len
;++i
)b
[i
]=(b
[i
]+b
[i
])%MOD
;for(int i
=0;i
<len
;++i
)b
[i
]=(b
[i
]+MOD
-d
[i
])%MOD
;
}
int n
,a
[MAX
],b
[MAX
];
int main()
{while(n
=read()){for(int i
=1;i
<=n
;++i
)a
[i
]=read()%MOD
;for(int i
=1;i
<=n
;++i
)a
[i
]=(MOD
-a
[i
])%MOD
;a
[0]++;int N
;for(N
=1;N
<=n
;N
<<=1);Inv(a
,b
,N
);printf("%d\n",b
[n
]);memset(a
,0,sizeof(a
));memset(b
,0,sizeof(b
));}return 0;
}
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