HDU5730 FFT+CDQ分治

题意：dp[n] = ∑ ( dp[n-i]*a[i] )+a[n], ( 1 <= i < n)

cdq分治。

计算出dp[l ~ mid]后，dp[l ~ mid]与a[1 ~ r-l]做卷积运算。

#include <bits/stdc++.h>
using namespace std;
const int N = 4e5+5;
const int mod = 313;
const double pi = acos(-1.0);
struct comp{
    double r,i;comp(double _r=0,double _i=0){r=_r;i=_i;}
    comp operator+(const comp x){return comp(r+x.r,i+x.i);}
    comp operator-(const comp x){return comp(r-x.r,i-x.i);}
    comp operator*(const comp x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
}x[N], y[N];
void FFT(comp a[],int n,int t){
    for(int i=1,j=0;i<n-1;i++){
        for(int s=n;j^=s>>=1,~j&s;);
        if(i<j)swap(a[i],a[j]);
    }
    for(int d=0;(1<<d)<n;d++){
        int m=1<<d,m2=m<<1;
        double o=pi/m*t;comp _w(cos(o),sin(o));
        for(int i=0;i<n;i+=m2){
            comp w(1,0);
            for(int j=0;j<m;j++){
                comp &A=a[i+j+m],&B=a[i+j],t=w*A;
                A=B-t;B=B+t;w=w*_w;
            }
        }
    }
    if(t==-1)for(int i=0;i<n;i++)a[i].r/=n;
}

int a[N], dp[N], n;
void cdq(int l,int r){
    if(l == r){
        dp[l] = (dp[l]+a[l])%mod;
        return;
    }
    int mid = l+r >> 1;
    cdq(l, mid);
    int len = 1;
    while(len <= (r-l+1)) len <<= 1;
    for(int i = 0; i < len; i++)
        x[i] = y[i] = comp(0, 0);
    for(int i = l; i <= mid; i++)
        x[i-l] = comp(dp[i], 0);
    for(int i = 1; l+i <= r; i++)
        y[i-1] = comp(a[i], 0);
    FFT(x, len, 1); FFT(y, len, 1);
    for(int i = 0; i < len; i++)
        x[i] = x[i]*y[i];
    FFT(x, len, -1);

    for(int i = mid+1; i <= r; i++){
        dp[i] += x[i-l-1].r+0.5;
        dp[i] %= mod;
    }
    cdq(mid+1,r);
}

int main(){
    while(scanf("%d", &n), n){
        for(int i = 1; i <= n; i++){
            scanf("%d", a+i);
            a[i] %= mod;
            dp[i] = 0;
        }
        cdq(1, n);
        printf("%d
", dp[n]);
    }
    return 0;
}

补：

因为做多项式逆运算在分治FFT之前，故做此题时首先有了一个多项式求逆的方法。

观察 dp[n] = ∑ ( dp[n-i]*a[i] )+a[n], ( 1 <= i < n) 

令a[0] = -1, 则 ∑ ( dp[n-i]*a[i] )+a[n] = 0, ( 0 <= i < n) 对任意n > 0成立

令dp[0] = 1

则dp序列与a序列卷积 = dp[0]*a[0]  +  ∑ ( dp[i-j]*a[j] ) xⁱ = dp[0]*a[0] = -1,

已知a序列，则dp序列为a的逆多项式序列*(-1)。

因为模313，最终各个系数 <= n(m-1)*(m-1) = 9734400000,  故选了一个大素数206158430209

写了一个多项式求逆的代码，不过不知什么原因WA了。

#include<cstdio>
#include<iostream>
typedef long long ll;
using namespace std;
const int N = 262144, K = 17;
ll n, m, i, k;
ll a[N+10], b[N+10], tmp[N+10], tmp2[N+10];
ll P = 206158430209LL, G = 22, g[K+1], ng[K+10], inv[N+10], inv2;
//ll P = 998244353, G = 3, g[K+1], ng[K+10], inv[N+10], inv2;
////////////
ll mul(ll a, ll b){
    ll ans = 0;
    while(b){
        if(b&1) ans += a;
        if(ans >= P) ans -= P;
        a = a+a;
        if(a >= P) a -= P;
        b >>= 1;
    }
    return ans;
}
ll pow(ll a, ll n){
    ll ans = 1;
    while(n){
        if(n&1)
            ans = mul(ans, a);
        a = mul(a, a);
        n >>= 1;
    }
    return ans;
}
////////////
void NTT(ll*a,int n,int t){
    for(int i=1,j=0;i<n-1;i++){
        for(int s=n;j^=s>>=1,~j&s;);
        if(i<j)swap(a[i], a[j]);
    }
    for(int d=0;(1<<d)<n;d++){
        int m=1<<d,m2=m<<1;
        ll _w=t==1?g[d]:ng[d];///////////////////////////////////

        for(int i=0;i<n;i+=m2)for(ll w=1,j=0;j<m;j++){
            ll&A=a[i+j+m],&B=a[i+j],t=mul(w, A);////////////(ll)w*A%P;
            A=B-t;if(A<0)A+=P;
            B=B+t;if(B>=P)B-=P;
            w=mul(w, _w);/////////(ll)w*_w%P;/////////////////////
        }
    }
    //if(t==-1)for(int i=0,j=inv[n];i<n;i++)a[i]=mul(a[i], j);///////(ll)a[i]*j%P;
    if(t==-1){
        ll j = pow(n, P-2);
        for(int i=0;i<n;i++)a[i]=mul(a[i], j);
    }
}
//给定a,求a的逆元b
void getinv(ll*a,ll*b,int n){
    if(n==1){b[0]=pow(a[0],P-2);return;}
    getinv(a,b,n>>1);
    int k=n<<1,i;
    for(i=0;i<n;i++)tmp[i]=a[i];
    for(i=n;i<k;i++)tmp[i]=b[i]=0;
    NTT(tmp,k,1),NTT(b,k,1);

    for(i=0;i<k;i++){
        ll xx = b[i];
        b[i] = mul(xx, (P+2-mul(xx, tmp[i]))%P);
    //b[i]=(ll)b[i]*(2-(ll)tmp[i]*b[i]%P)%P;
    //if(b[i]<0)b[i]+=P;
    }
    NTT(b,k,-1);
    for(i=n;i<k;i++)b[i]=0;
}
int main(){
    for(g[K]=pow(G,(P-1)/N),ng[K]=pow(g[K],P-2),i=K-1;~i;i--){
        //g[i]=(ll)g[i+1]*g[i+1]%P,ng[i]=(ll)ng[i+1]*ng[i+1]%P;
        g[i] = mul(g[i+1], g[i+1]);
        ng[i] = mul(ng[i+1], ng[i+1]);
    }
    //for(inv[1]=1,i=2;i<=N;i++)inv[i]=(ll)(P-inv[P%i])*(P/i)%P;inv2=inv[2];
    while(scanf("%lld", &n), n){
        int len = 1;
        while(len <= n) len <<= 1;
        //len <= 131072
        a[0] = (P-1)%313;
        for(i = 1; i <= n; i++){
            scanf("%lld", a+i);
            a[i] %= 313;
        }
        for(i = n+1; i < len; i++) a[i] = 0;

        getinv(a, b, len);
        b[n] = b[n]%313*(P-1LL)%313;
//        for(i = 0; i < len; i++)
//            b[i] = mul(b[i], P-1);
//              b[i] = b[i]*(P-1LL)%P;
        printf("%lld
", b[n]);
    }
    return 0;
}