hdu5322 Hope

　　设dp[n]为n个数字排列时候的答案，那么可以得到dp方程

　　dp[n]=Σdp[n-i]*c(n-1,i-1)*(i-1)!*i^2(1<=i<=n)

　　然后上式可以化成卷积形式，分治FFT即可。复杂度O(nlogn^2)

　　

　　代码

#include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    typedef long long LL;
    const int P = *(<<)+;
    const int N =  << ;
    const int G = ;
    const int NUM = ;

    LL  wn[NUM],cnt[N],jc[N],dp[N],dp2[N],inv[N];
    LL  a[N], b[N];
    char A[N], B[N];
    int len,i;
    LL quick_mod(LL a, LL b, LL m)
    {
        LL ans = ;
        a %= m;
        while(b)
        {
            if(b & )
            {
                ans = ans * a % m;
                b--;
            }
            b >>= ;
            a = a * a % m;
        }
        return ans;
    }

    void GetWn()
    {
        for(int i=; i<NUM; i++)
        {
            int t =  << i;
            wn[i] = quick_mod(G, (P - ) / t, P);
        }
    }
    void Rader(LL a[], int len)
    {
        int j = len >> ;
        for(int i=; i<len-; i++)
        {
            if(i < j) swap(a[i], a[j]);
            int k = len >> ;
            while(j >= k)
            {
                j -= k;
                k >>= ;
            }
            if(j < k) j += k;
        }
    }

    void NTT(LL a[], int len, int on)
    {
        Rader(a, len);
        int id = ;
        for(int h = ; h <= len; h <<= )
        {
            id++;
            for(int j = ; j < len; j += h)
            {
                LL w = ;
                for(int k = j; k < j + h / ; k++)
                {
                    LL u = a[k] % P;
                    LL t = w * (a[k + h / ] % P) % P;
                    a[k] = (u + t) % P;
                    a[k + h / ] = ((u - t) % P + P) % P;
                    w = w * wn[id] % P;
                }
            }
        }
        if(on == -)
        {
            for(int i = ; i < len / ; i++)
                swap(a[i], a[len - i]);
            LL Inv = quick_mod(len, P - , P);
            for(int i = ; i < len; i++)
                a[i] = a[i] % P * Inv % P;
        }
    }
    void Conv(LL a[], LL b[], int n)
    {
        NTT(a, n, );
        NTT(b, n, );
        for(int i = ; i < n; i++)
            a[i] = a[i] * b[i] % P;
        NTT(a, n, -);
    }
    void solve(long long l,long long r)
    {
        int m,i;
        if (l==r)
        {
            dp[l]=(dp[l]+jc[l-]*l%P*l%P)%P;
            return;
        }
        m=(l+r)>>;
        solve(l,m);

        len=;
        while (len<=r-l+) len<<=;
        for (i=l;i<=r;i++)
        {
            if (i<=m)
                a[i-l]=dp[i]*inv[i]%P;
            else
                a[i-l]=;

            if (i<r)
            b[i-l+]=inv[i-l]*jc[i-l]%P*(i-l+)%P*(i-l+)%P;
            else
            b[i-l+]=;
        }
        for (i=r-l+;i<=len;i++)
        {
            a[i]=;b[i]=;
        }
         b[]=;

        Conv(a,b,len);

        for (i=m+;i<=r;i++)
        {
            if (i->)
            dp[i]=(dp[i]+a[i-l]*jc[i-])%P;
        }

        solve(m+,r);
    }
    int main()
    {
        GetWn();
        jc[]=;inv[]=;
        for (i=;i<=;i++)
        {
            jc[i]=jc[i-]*i %P;
            inv[i]=quick_mod(jc[i],P-,P);
        }
        solve(,);
        int n;
        while (scanf("%d",&n)==)
        printf("%I64d\n",dp[n]);
    }