BZOJ 2111 排列计数

f[i]=f[l]*f[r]*C(size[l]+size[r],size[l]).

需要lucas.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 1000050
using namespace std;
long long n,mod,inv1[maxn],inv2[maxn],f[maxn],size[maxn];
long long f_pow(long long x,long long y)
{
    long long ans=,base=x;
    while (y)
    {
        if (y&) ans=(ans*base)%mod;
        base=(base*base)%mod;
        y>>=;
    }
    return ans;
}
void get_table()
{
    inv1[]=;inv2[]=mod+;
    for (long long i=;i<=n;i++)
    {
        inv1[i]=inv1[i-]*i%mod;
        inv2[i]=f_pow(inv1[i],mod-);
    }
}
long long comb(long long n,long long m)
{
    return inv1[n]*inv2[m]%mod*inv2[n-m]%mod;
}
long long lucas(long long n,long long m)
{
    if (!m) return ;
    return comb(n%mod,m%mod)*lucas(n/mod,m/mod)%mod;
}
int main()
{
    scanf("%lld%lld",&n,&mod);
    get_table();
    for (long long i=n;i>=;i--)
    {
        size[i]=;long long l=,r=;f[i]=;
        if ((i<<)<=n) {size[i]+=size[i<<];l=size[i<<];f[i]=f[i]*f[i<<]%mod;}
        if ((i<<|)<=n) {size[i]+=size[i<<|];r=size[i<<|];f[i]=f[i]*f[i<<|]%mod;}
        f[i]=f[i]*lucas(l+r,l)%mod;
    }
    printf("%lld\n",f[]%mod);
    return ;
}