bzoj 2111: [ZJOI2010]Perm 排列计数 （dp+卢卡斯定理）

Posted 2020-09-29 GraceSkyer

bzoj 2111: [ZJOI2010]Perm 排列计数

1 ≤ N ≤ 10^6, P≤ 10^9

题意：求1~N的排列有多少种小根堆

   1: #include<cstdio>

   2: using namespace std;

   3: const int N = 1e6+5;

   4: typedef long long LL;

   5: LL m, p, T, x, y, F[N];

   6: LL n, size[N<<1];

   7: LL f[N];

   8: LL inv(LL t, LL p) {

   9:     return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;

  10: }

  11: LL C(LL n, LL m){

  12:     if(n < m) return 0;

  13:     if(n < p && m < p)

  14:         return F[n] * 1ll * inv(F[n-m]%p, p) % p * inv(F[m]%p, p) % p;

  15:     return C(n/p, m/p) * C(n%p, m%p) %p;

  16: }

  17: int main(){

  18:     int i;

  19:     scanf("%d%lld", &n, &p);

  20:     F[0] = 1;

  21:     for(i = 1; i <= n&&i < p; i++) F[i] = F[i-1] * i %p;//阶乘预处理

  22:     for(i = n; i; i--) {

  23:         size[i] = size[i<<1] + size[i<<1|1] + 1;

  24:         f[i] = C(size[i]-1, size[i<<1])*((i<<1)>n?1:f[i<<1])%p*((i<<1|1)>n?1:f[i<<1|1])%p;

  25:     }

  26:     printf("%lld\n", f[1]);

  27: }