SCL--FFT & NTT

2015-07-28 02:26:59

总结:

(1)关于FFT,算法导论讲得非常好,证明详细,课后题很有启发性。这里先上花了一天多时间综合总结起来的模板,效率还可。

  大致流程有倍次,求值,乘法,插值。求值点是 n 次单位复数根。

(2)关于NTT,推荐一下 Acdreamer 的 blog ,有关原根的知识:blog

  然后就是FFT整数域中模运算下的形式。求值点变成了模数P的原根的次幂,g^((P-1)/t),t = 2^k

  模数的取法:(479 << 21) + 1 or 998244353(2^23 * 7 * 17),其原根为3 。

 

 

以UOJ的多项式乘法为测试平台

NTT:

const int P = (479 << 21) + 1; //费马素数
const int G = 3; //原根
const int MAXN = (1 << 18) + 10;
const int NUM = 20;

int rev[MAXN];
int A1[MAXN],A2[MAXN],wn[2][NUM];
int n,m,n3,N,bit;

int Q_pow(int x,int y,int mod){
    int res = 1;
    x %= mod;
    while(y){
        if(y & 1) res = 1ll * res * x % mod;
        x = 1ll * x * x % mod;
        y >>= 1;
    }
    return res;
}

void Pre_cal(){
    n3 = n + m - 1; //结果多项式的次数界
    memset(rev,0,sizeof(rev));
    for(N = 1,bit = 0; N < n3; N <<= 1,++bit); //DFT底层
    for(int i = 1; i < N; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    for(int i = 0; i < NUM; ++i){
        int t = 1 << i;
        wn[0][i] = Q_pow(G,(P - 1) / t,P); //预处理求值点
        wn[1][i] = Q_pow(wn[0][i],P - 2,P); //求值点逆元
    }
}

void NTT(int *A,int n,int f){
    for(int i = 0; i < n; ++i) if(i < rev[i]) swap(A[i],A[rev[i]]);
    int id = (f == -1) ? 1 : 0,p = 1;
    for(int m = 2; m <= n; m <<= 1,++p){ //m次单位根
        for(int k = 0; k < n; k += m){ //遍历每一块
            for(int j = k,w = 1; j < k + (m >> 1); ++j){ //折半
                int t = 1ll * w * A[j + (m >> 1)] % P; //右项
                int u = A[j] % P; //左项 (此处取模待商讨)
                if((A[j] = u + t) >= P) A[j] -= P;
                if((A[j + (m >> 1)] = u - t) < 0) A[j + (m >> 1)] += P;
                w = 1ll * w * wn[id][p] % P;
            }
        }
    }
    if(f == -1){
        int inv = Q_pow(n,P - 2,P);
        for(int i = 0; i < n; ++i) A[i] = 1ll * A[i] * inv % P;
    }
}

void Solve(){
    //流程:倍次,求值,乘法,插值
    Pre_cal();
    NTT(A1,N,1); //求值
    NTT(A2,N,1);
    for(int i = 0; i < N; ++i) A1[i] = 1ll * A1[i] * A2[i] % P; //注意ll
    NTT(A1,N,-1); //插值
    for(int i = 0; i < n3 - 1; ++i)  printf("%d ",A1[i]);
    printf("%d\n",A1[n3 - 1]);
}

int main(){
    scanf("%d%d",&n,&m);
    n++,m++; //次数界
    for(int i = 0; i < n; ++i) scanf("%d",A1 + i);
    for(int i = 0; i < m; ++i) scanf("%d",A2 + i);
    Solve();
    return 0;
}

 

FFT:

const int MAXN = (1 << 18) + 10;
const double DPI = 2.0 * acos(-1.0);

int n,m,n3,N,bit;
int rev[MAXN];

struct CP{ //复数类
    double a,b;
    CP(double ta = 0,double tb = 0) : a(ta) , b(tb) {}
}A1[MAXN],A2[MAXN];

inline CP operator * (CP &a,CP &b){return CP(a.a*b.a-a.b*b.b,a.a*b.b+a.b*b.a);}
inline CP operator + (CP &a,CP &b){return CP(a.a + b.a,a.b + b.b);}
inline CP operator - (CP &a,CP &b){return CP(a.a - b.a,a.b - b.b);}

void Pre_cal(){
    n3 = n + m - 1; //结果多项式次数界
    memset(rev,0,sizeof(rev));
    for(N = 1,bit = 0; N < n3; N <<= 1,++bit);
    for(int i = 1; i < N; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
}

void FFT(CP *A,int n,int f){
    for(int i = 0; i < n; ++i) if(i < rev[i]) swap(A[i],A[rev[i]]);
    for(int m = 2; m <= n; m <<= 1){ //DFT结果的项数
        CP wm(cos(DPI / m),f * sin(DPI / m)); //m次单位复数根
        for(int k = 0; k < n; k += m){ //遍历每一块
            CP w(1,0);
            for(int j = k; j < k + (m >> 1); ++j){ //折半,关键
                CP t = w * A[j + (m >> 1)]; //右项
                CP u = A[j]; //左项
                A[j] = u + t;
                A[j + (m >> 1)] = u - t;
                w = w * wm; //更新w
            }
        }
    }
    if(f == -1) for(int i = 0; i < n; ++i) A[i].a /= n;
}

void Solve(){
    //流程:倍次,求值,乘法,插值
    Pre_cal();
    FFT(A1,N,1); //求值1
    FFT(A2,N,1); //求值2
    for(int i = 0; i < N; ++i) A1[i] = A1[i] * A2[i];
    FFT(A1,N,-1); //插值
    for(int i = 0; i < n3 - 1; ++i) printf("%d ",(int)(A1[i].a + 0.5));
    printf("%d\n",(int)(A1[n3 - 1].a + 0.5));
}

int main(){
    scanf("%d%d",&n,&m);
    n++; m++;
    for(int i = 0; i < n; ++i) scanf("%lf",&A1[i].a),A1[i].b = 0;
    for(int i = 0; i < m; ++i) scanf("%lf",&A2[i].a),A2[i].b = 0;;
    Solve();
    return 0;
}

 

posted @ 2015-07-28 02:32  Naturain  阅读(380)  评论(0编辑  收藏  举报