生活随笔
收集整理的這篇文章主要介紹了
ACM入门之【组合数】
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
組合數模板一
const int N
= 2010 ;
const int mod
= 1e9 + 7 ;
int C
[ N
] [ N
] , t
, a
, b
;
void init ( )
{ for ( int i
= 0 ; i
< N
; i
++ ) { for ( int j
= 0 ; j
<= i
; j
++ ) { if ( j
== 0 ) C
[ i
] [ j
] = 1 ; else C
[ i
] [ j
] = ( C
[ i
- 1 ] [ j
- 1 ] + C
[ i
- 1 ] [ j
] ) % mod
; } }
}
組合數模板二
typedef long long int LL
;
const int N
= 1e6 + 10 ;
const int mod
= 1e9 + 7 ;
LL f
[ N
] , inf
[ N
] , n
;
LL
quick_mi ( LL a
, LL b
, LL p
)
{ LL sum
= 1 ; while ( b
) { if ( b
& 1 ) sum
= sum
* a
% p
; b
>>= 1 ; a
= a
* a
% p
; } return sum
% p
;
}
void init ( )
{ f
[ 0 ] = 1 , inf
[ 0 ] = 1 ; for ( int i
= 1 ; i
< N
; i
++ ) { f
[ i
] = f
[ i
- 1 ] * i
% mod
; inf
[ i
] = inf
[ i
- 1 ] * quick_mi ( i
, mod
- 2 , mod
) % mod
; }
}
LL
query ( LL a
, LL b
)
{ return f
[ a
] * inf
[ a
- b
] % mod
* inf
[ b
] % mod
;
}
組合數模板三
typedef long long int LL
;
LL
qmi ( LL a
, LL b
, LL p
)
{ LL res
= 1 ; while ( b
) { if ( b
& 1 ) res
= res
* a
% p
; a
= a
* a
% p
; b
= b
>> 1 ; } return res
;
}
LL
C ( LL a
, LL b
, LL p
)
{ if ( b
> a
) return 0 ; if ( b
> a
- b
) b
= a
- b
; LL x
= 1 , y
= 1 ; for ( int i
= 0 ; i
< b
; i
++ ) { x
= x
* ( a
- i
) % p
; y
= y
* ( i
+ 1 ) % p
; } return x
* qmi ( y
, p
- 2 , p
) % p
;
}
LL
lucas ( LL a
, LL b
, LL p
)
{ if ( a
< p
&& b
< p
) return C ( a
, b
, p
) ; return C ( a
% p
, b
% p
, p
) * lucas ( a
/ p
, b
/ p
, p
) % p
;
}
組合數模板四
#include <bits/stdc++.h>
using namespace std
;
const int N
= 5010 ;
int sum
[ N
] , prime
[ N
] , a
, b
, cnt
;
bool st
[ N
] ;
void init ( int n
)
{ for ( int i
= 2 ; i
<= n
; i
++ ) { if ( ! st
[ i
] ) prime
[ cnt
++ ] = i
; for ( int j
= 0 ; prime
[ j
] <= n
/ i
; j
++ ) { st
[ prime
[ j
] * i
] = true ; if ( i
% prime
[ j
] == 0 ) break ; } }
}
int get ( int n
, int p
)
{ int res
= 0 ; while ( n
) res
+ = n
/ p
, n
= n
/ p
; return res
;
}
vector
< int > mul ( vector
< int > A
, int b
)
{ vector
< int > C
; int t
= 0 ; for ( int i
= 0 ; i
< A
. size ( ) || t
; i
++ ) { if ( i
< A
. size ( ) ) t
+ = A
[ i
] * b
; C
. push_back ( t
% 10 ) ; t
/ = 10 ; } while ( C
. size ( ) > 1 && C
. back ( ) == 0 ) C
. pop_back ( ) ; return C
;
}
vector
< int > ans
;
int main ( void )
{ cin
>> a
>> b
; init ( a
) ; ans
. push_back ( 1 ) ; for ( int i
= 0 ; i
< cnt
; i
++ ) sum
[ i
] = get ( a
, prime
[ i
] ) - get ( a
- b
, prime
[ i
] ) - get ( b
, prime
[ i
] ) ; for ( int i
= 0 ; i
< cnt
; i
++ ) { for ( int j
= 0 ; j
< sum
[ i
] ; j
++ ) ans
= mul ( ans
, prime
[ i
] ) ; } for ( int i
= ans
. size ( ) - 1 ; i
>= 0 ; i
-- ) cout
<< ans
[ i
] ; return 0 ;
}
卡特蘭數
typedef long long int LL
;
LL f
[ 25 ] , n
;
LL
get ( int n
)
{ f
[ 0 ] = 1 ; for ( int i
= 1 ; i
<= n
; i
++ ) f
[ i
] = f
[ i
- 1 ] * ( 4 * i
- 2 ) / ( i
+ 1 ) ; return f
[ n
] ;
}
885. 求組合數 I
總結
以上是生活随笔 為你收集整理的ACM入门之【组合数】 的全部內容，希望文章能夠幫你解決所遇到的問題。
如果覺得生活随笔 網站內容還不錯，歡迎將生活随笔 推薦給好友。
