bzoj 2111: [ZJOI2010]Perm 排列计数

神题。。。

扒自某神犇题解：

http://blog.csdn.net/aarongzk/article/details/50655471

#include<bits/stdc++.h>
#define LL long long
using namespace std;
inline LL ra()
{
    LL x=0; char ch=getchar();
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x;
}

const int maxn=2000005;

LL n,mod,f[maxn],size[maxn];
LL fac[maxn],inv[maxn];

LL C(int n, int m)
{
    if (n<m) return 0;
    if (n<mod && m<mod) return fac[n]*inv[m]%mod*inv[n-m]%mod;
    return C(n/mod,m/mod)*C(n%mod,m%mod);
}

int main()
{
    n=ra(); mod=ra(); fac[0]=1;
    for (int i=1; i<=n; i++) fac[i]=fac[i-1]*i%mod;
    inv[0]=inv[1]=1;
    for (int i=2; i<=n; i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    for (int i=2; i<=n; i++) inv[i]=inv[i]*inv[i-1]%mod;
    for (int i=n; i>=1; i--)
    {
        size[i]=size[i<<1]+size[i<<1|1]+1;
        f[i]=((i<<1)>n?1:f[i<<1])*((i<<1|1)>n?1:f[i<<1|1])%mod*C(size[i]-1,size[i<<1])%mod;
    }    
    cout<<f[1]<<endl;
    return 0;
}