題目大意：有n個數(shù)字一開始全0，每次隨機一個數(shù)字++，問期望多少步后第一次有個位置的數(shù)字的值是k。n≤50,k≤1000n\le50,k\le1000n≤50,k≤1000。
題解：
顯然k=1是個min-max容斥。因此min-max容斥。假設枚舉的集合大小是aaa。
一開始場上的做法是，期望轉為∑i≥1P(ans≥i)\sum_{i\ge1}P(ans\ge i)∑i≥1?P(ans≥i)，然后P(ans≥i)P(ans\ge i)P(ans≥i)就是說到i?1i-1i?1的時候還沒好的概率，然后再枚舉這i-1次操作中多少次落在左半邊，整理式子后發(fā)現(xiàn)有個組合數(shù)數(shù)列點積等比數(shù)列的求和，總之推導一波后發(fā)現(xiàn)要計算大小為a的集合i步后不存在>=k的數(shù)值的方案數(shù)，發(fā)現(xiàn)這玩意只能n2k2n^2k^2n2k2dp，但是是卷積，場上寫個NTT過了。
另一種做法是直接枚舉：∑i≥1P(ans=i)i\sum_{i\ge 1}P(ans=i)i∑i≥1?P(ans=i)i，然后還是枚舉左邊的次數(shù)j，右邊還是組合數(shù)數(shù)列點積等比數(shù)列的求和（其實算出來就是na\frac naan?），推導一下發(fā)現(xiàn)要算大小為a的集合i步后恰好有個位置是k的方案數(shù)，這個就用欽定某個數(shù)第i步是出現(xiàn)恰好k次的總方案數(shù)減去不合法的位置。后者用個前綴和即可。其實就是推式子很麻煩，但直接冷靜思考一波就是a的答案乘以na\frac naan?的比率即可。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
#define lint long long
#define p 998244353
#define ull unsigned lint
#define db long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define gc getchar()
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std
;
typedef pair
<int,int> pii
;
typedef set
<int>::iterator sit
;
inline int inn()
{int x
,ch
;while((ch
=gc
)<'0'||ch
>'9');x
=ch
^'0';while((ch
=gc
)>='0'&&ch
<='9')x
=(x
<<1)+(x
<<3)+(ch
^'0');return x
;
}
const int N
=52,K
=1002;
int f
[2][N
*K
],fac
[N
*K
],facinv
[N
*K
],inv
[N
*K
],mi1
[N
*K
],mi2
[N
*K
];
inline int fast_pow(int x
,int k
,int ans
=1) { for(;k
;k
>>=1,x
=(lint
)x
*x
%p
) (k
&1)?ans
=(lint
)ans
*x
%p
:0;return ans
; }
inline int C(int n
,int m
) { return assert(n
>=m
&&m
>=0),(lint
)fac
[n
]*facinv
[m
]%p
*facinv
[n
-m
]%p
; }
inline int prelude(int n
)
{rep(i
,fac
[0]=1,n
) fac
[i
]=(lint
)i
*fac
[i
-1]%p
;facinv
[n
]=fast_pow(fac
[n
],p
-2);for(int i
=n
-1;i
>=0;i
--) facinv
[i
]=(i
+1ll)*facinv
[i
+1]%p
;rep(i
,1,n
) inv
[i
]=(lint
)fac
[i
-1]*facinv
[i
]%p
;return 0;
}
int main()
{int n
=inn(),k
=inn(),ans
=0;prelude(n
*k
);rep(i
,1,n
){int *now
=f
[i
&1],*pre
=f
[(i
-1)&1],s
=0;rep(j
,k
,(i
-1)*(k
-1)+1) pre
[j
]=(pre
[j
]+(i
-1ll)*pre
[j
-1])%p
;rep(j
,mi1
[0]=1,i
*(k
-1)+1) mi1
[j
]=mi1
[j
-1]*(i
-1ll)%p
;rep(j
,mi2
[0]=1,i
*(k
-1)+2) mi2
[j
]=(lint
)mi2
[j
-1]*inv
[i
]%p
;rep(j
,k
,i
*(k
-1)+1)now
[j
]=(lint
)i
*C(j
-1,k
-1)%p
*(mi1
[j
-k
]-pre
[j
-k
])%p
,(now
[j
]<0?now
[j
]+=p
:0),s
=(s
+(lint
)now
[j
]*j
%p
*mi2
[j
+1])%p
;if(i
&1) ans
=(ans
+(lint
)C(n
,i
)*s
)%p
;else ans
=(ans
-(lint
)C(n
,i
)*s
)%p
,(ans
<0?ans
+=p
:0);}return !printf("%lld\n",(lint
)ans
*n
%p
);
}
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define Rep(i,v) rep(i,0,(int)v.size()-1)
#define lint long long
#define p 998244353
#define ull unsigned lint
#define db long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define gc getchar()
#define debug(x) cerr<<#x<<"="<<x
#define sp <<" "
#define ln <<endl
using namespace std
;
typedef pair
<int,int> pii
;
typedef set
<int>::iterator sit
;
inline int inn()
{int x
,ch
;while((ch
=gc
)<'0'||ch
>'9');x
=ch
^'0';while((ch
=gc
)>='0'&&ch
<='9')x
=(x
<<1)+(x
<<3)+(ch
^'0');return x
;
}
const int N
=52,K
=1005,QWQ
=67000;
int dp
[N
][QWQ
],fac
[N
*K
],facinv
[N
*K
],mi
[N
*K
],tmp
[QWQ
];
inline int fast_pow(int x
,int k
,int ans
=1) { for(;k
;k
>>=1,x
=(lint
)x
*x
%p
) (k
&1)?ans
=(lint
)ans
*x
%p
:0;return ans
; }
inline int sol(int x
,int s
) { return (s
&1)?(x
?p
-x
:0):x
; }
inline int C(int n
,int m
) { return (lint
)fac
[n
]*facinv
[m
]%p
*facinv
[n
-m
]%p
; }
namespace NTT_space
{const int N
=67000;int r
[N
],*dwg
[N
],*dwgi
[N
];inline int prelude_dwg(){int n
=N
-2;for(int i
=2,t
=1;i
<=n
;i
<<=1,t
++){dwg
[t
]=new
int[i
],dwgi
[t
]=new
int[i
];int *d
=dwg
[t
],*di
=dwgi
[t
];d
[0]=di
[0]=1;int w
=fast_pow(3,(p
-1)/i
),wi
=fast_pow(3,p
-1-(p
-1)/i
);rep(j
,1,i
-1) d
[j
]=(lint
)d
[j
-1]*w
%p
,di
[j
]=(lint
)di
[j
-1]*wi
%p
;}return 0;}inline int pre(int m
){int n
=1,L
=0;while(n
<=m
) n
<<=1,L
++;rep(i
,1,n
-1) r
[i
]=(r
[i
>>1]>>1)|((i
&1)<<(L
-1));return n
;}inline int NTT(int *a
,int n
,int s
){rep(i
,1,n
-1) if(i
<r
[i
]) swap(a
[i
],a
[r
[i
]]);for(int i
=2,c
=1;i
<=n
;i
<<=1,c
++){int *d
=(s
>0?dwg
[c
]:dwgi
[c
]);for(int j
=0,t
=i
>>1,x
,y
;j
<n
;j
+=i
) rep(k
,0,t
-1)x
=a
[j
+k
],y
=(lint
)d
[k
]*a
[j
+k
+t
]%p
,a
[j
+k
]=x
+y
,(a
[j
+k
]>=p
?a
[j
+k
]-=p
:0),a
[j
+k
+t
]=x
-y
,(a
[j
+k
+t
]<0?a
[j
+k
+t
]+=p
:0);}if(s
<0) for(int i
=0,v
=fast_pow(n
,p
-2);i
<n
;i
++) a
[i
]=(lint
)a
[i
]*v
%p
;return 0;}
}using NTT_space
::NTT
;
inline int prelude(int n
,int k
)
{dp
[0][0]=1;int m
=max(n
,n
*(k
-1));rep(i
,fac
[0]=1,m
) fac
[i
]=(lint
)fac
[i
-1]*i
%p
;facinv
[m
]=fast_pow(fac
[m
],p
-2);for(int i
=m
-1;i
>=0;i
--) facinv
[i
]=(i
+1ll)*facinv
[i
+1]%p
;int t
=NTT_space
::pre(n
*(k
-1));NTT(dp
[0],t
,1);rep(i
,0,k
-1) tmp
[i
]=facinv
[i
];NTT(tmp
,t
,1);rep(i
,1,n
) rep(j
,0,t
-1) dp
[i
][j
]=(lint
)dp
[i
-1][j
]*tmp
[j
]%p
;rep(i
,1,n
) NTT(dp
[i
],t
,-1);rep(i
,1,n
) rep(j
,0,i
*(k
-1)) dp
[i
][j
]=(lint
)dp
[i
][j
]*fac
[j
]%p
;return 0;
}
int main()
{NTT_space
::prelude_dwg();int n
=inn(),k
=inn(),ans
=0;prelude(n
,k
);rep(i
,1,n
){int s
=0,v
=fast_pow(i
,p
-2);rep(j
,mi
[0]=1,i
*(k
-1)+1)mi
[j
]=(lint
)mi
[j
-1]*v
%p
;rep(j
,0,i
*(k
-1))s
=(s
+(lint
)mi
[j
+1]*dp
[i
][j
])%p
;ans
+=sol((lint
)C(n
,i
)*s
%p
,i
+1);if(ans
>=p
) ans
-=p
;}return !printf("%lld\n",(lint
)ans
*n
%p
);
}
總結
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