傳送門
題意簡述：現(xiàn)在有一些號碼由000~999中的某些數(shù)字組成（會給出），號碼總長度為nnn，問有多少個號碼滿足前n2\frac n22n?個數(shù)碼的和等于后n2\frac n22n?個數(shù)碼的和（保證nnn是偶數(shù)），答案對998244353998244353998244353取模。
思路：
一道挺顯然的生成函數(shù)+快速冪。
考慮到前n2\frac n22n?個數(shù)碼和的生成函數(shù)和后n2\frac n22n?個數(shù)碼和的生成函數(shù)是相同的，因此直接求出前n2\frac n22n?個數(shù)碼和的生成函數(shù)，然后對于每一項的系數(shù)平方加起來即可。
代碼：
#include<bits/stdc++.h>
#define ri register int
using namespace std
;
const int mod
=998244353,N
=2e5+5;
typedef long long ll
;
int n
,k
,tim
,lim
;
vector
<int>A
,B
,pos
;
inline void init(const int&up
){lim
=1,tim
=0;while(lim
<=up
)lim
<<=1,++tim
;pos
.resize(lim
),A
.resize(lim
),B
.resize(lim
);for(ri i
=0;i
<lim
;++i
)pos
[i
]=(pos
[i
>>1]>>1)|((i
&1)<<(tim
-1));
}
inline int add(int a
,int b
){return a
+b
>=mod
?a
+b
-mod
:a
+b
;}
inline int dec(int a
,int b
){return a
>=b
?a
-b
:a
-b
+mod
;}
inline int mul(int a
,int b
){return (ll
)a
*b
%mod
;}
inline int ksm(int a
,int p
){int ret
=1;for(;p
;p
>>=1,a
=mul(a
,a
))if(p
&1)ret
=mul(ret
,a
);return ret
;}
inline void ntt(vector
<int>&a
,const int&type
){for(ri i
=0;i
<lim
;++i
)if(i
<pos
[i
])swap(a
[i
],a
[pos
[i
]]);for(ri w
,wn
,typ
=type
==1?3:(mod
+1)/3,mult
=(mod
-1)/2,mid
=1;mid
<lim
;mid
<<=1,mult
>>=1){wn
=ksm(typ
,mult
);for(ri j
=0,len
=mid
<<1;j
<lim
;j
+=len
)for(ri w
=1,a0
,a1
,k
=0;k
<mid
;++k
,w
=mul(w
,wn
)){a0
=a
[j
+k
],a1
=mul(w
,a
[j
+k
+mid
]);a
[j
+k
]=add(a0
,a1
),a
[j
+k
+mid
]=dec(a0
,a1
);}}if(type
==-1)for(ri i
=0,inv
=ksm(lim
,mod
-2);i
<lim
;++i
)a
[i
]=mul(a
[i
],inv
);
}
struct poly
{vector
<int>a
;poly(int k
=0,int x
=0){a
.resize(k
+1),a
[k
]=x
;}inline int&operator[](const int&k
){return a
[k
];}inline const int&operator[](const int&k
)const{return a
[k
];}inline int deg()const{return a
.size()-1;}inline poly
extend(int k
){poly ret
=*this;return ret
.a
.resize(k
+1),ret
;}friend inline poly
operator^(const poly
&a
,const int&k
){init(a
.deg()*k
);for(ri i
=0;i
<=a
.deg();++i
)A
[i
]=B
[i
]=a
[i
];for(ri i
=a
.deg()+1;i
<lim
;++i
)A
[i
]=B
[i
]=0;int p
=k
-1;ntt(A
,1),ntt(B
,1);while(p
){if(p
&1)for(ri i
=0;i
<lim
;++i
)B
[i
]=mul(B
[i
],A
[i
]);for(ri i
=0;i
<lim
;++i
)A
[i
]=mul(A
[i
],A
[i
]);p
>>=1;}poly ret
;return ntt(B
,-1),ret
.a
=B
,ret
.extend(a
.deg()*k
);}
};
int main(){poly
a(9);int mx
=0;scanf("%d%d",&n
,&k
),n
>>=1;for(ri i
=1,v
;i
<=k
;++i
)scanf("%d",&v
),a
[v
]=1,mx
=max(mx
,v
);a
=a
.extend(mx
);a
=(a
^n
);int ans
=0;for(ri i
=0;i
<=mx
*n
;++i
)ans
=add(ans
,mul(a
[i
],a
[i
]));cout
<<ans
;return 0;
}
轉(zhuǎn)載于:https://www.cnblogs.com/ldxcaicai/p/10367716.html
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