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多项式全家桶

时间：2019-12-08 10:48:22      阅读：103      评论：0      收藏：0      [点我收藏+]

标签：朴素   git   dft   离散傅里叶变换   lin   快速   傅里叶变换   cpp   lex   

多项式全家桶(更新至快速幂)

开始爆肝多项式

1. FFT快速傅里叶变换

流程: 将多项式\(\Theta (nlog_n)\)转成点值表示形式 进行卷积, 再\(\Theta (nlog_n)\) 转回来

离散傅里叶变换:

朴素转为点值, 需要将一个一个x带入, 而这里傅里叶搞到了几个可以优化的复数根

利用复数, 在复平面上画出一个单位圆, 将单位圆n等分, 每一个等分点为$\omega_n^i $,

接下来将\(\omega^1_n \omega^2_n\cdots\omega^n_n\) n个根带入, 其中\(\omega_n^i = cos(\frac{k}{n}2\pi) + i * sin(\frac{k}{n}2\pi)\)

稍等, 还要几个小性质才可以:

性质一: \(\omega^{2k}_{2n} = \omega_n^k\)

性质二: \(\omega_n^{k+\frac{n}{2}} = - \omega_n^k\)

接下来开始快速傅里叶变换:

设A(x) = \(a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}\)

利用分治, 将A按x的指数分为奇偶两部分

\(A(x)=(a_0+a_2x^2+\cdots + a_{n-2}x^{n-2})+(a_1x+a_3x^3+\cdots\) $ a_{n-1}x^{n-1})$

设偶部分 \(A_1(x) = a_0 + a_2x + a_4x^2 + \cdots + a_{n-2}x^{\frac{n}{2}-1}\)

奇部分\(A_2(x) = a_1 + a_3x + a_5x^2 + \cdots + a_{n-1}x^{\frac{n}{2}-1}\)

则有: \(A(x) = A_1(x^2) + x A_2(x^2)\)

设k < n/2 带入\(x = \omega^k_n\)

\(A(\omega^k_n) = A_1(\omega^{2k}_n) + \omega_n^kA_2(\omega^{2k}_n)\)

再带入 \(\omega^{k+\frac{n}{2}}_n\)

得\(A(\omega^{k+\frac{n}{2}}_n) = A_1(\omega^{2k}_n) -\omega_n^kA_2(\omega^{2k}_n)\)

你fa♂现了吗, 分治时, 两个求值一次解决

IDFT: 离散傅里叶逆变换

设\((y_0,y_1,\cdots,y_{n-1})\) 为多项式A(x) = \(a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}\)的离散傅里叶变换

设多项式 B(x) = A(x) = \(y_0 + y_1x + y_2x^2 + \cdots + y_{n-1}x^{n-1}\)

将n个单位根的倒数带入得新的离散傅里叶变换

此处略去推导, 得\(a_i = \frac{z_i}{n}\)

非递归版fft

每个数递归到底层时的二进制表示恰是原来的表示翻转得到的

如 6(011) 到 3(110)

再加上蝴蝶变换就可以啦

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 3000600;


template <typename T>
void read(T &x) {
    x = 0; bool f = 0;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    if (f) x = -x;
}

struct Complex {
    double x, y;
    Complex (double xx = 0,double yy = 0) {x=xx,y=yy;}
    Complex operator + (Complex &i) const {
        return Complex(x + i.x, y + i.y);
    }
    Complex operator - (Complex &i) const {
        return Complex(x - i.x, y - i.y);
    }
    Complex operator * (Complex &i) const {
        return Complex(x * i.x - y * i.y, x * i.y + y * i.x);
    }
}A[N], B[N];

const double Pi = acos(-1.0);

int n, m;

int lim = 1, L;
int r[N];

void FFT(Complex *a,int type) {
    for (int i = 0;i < lim; i++)
    if (r[i] > i) swap(a[r[i]], a[i]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        Complex T(cos(Pi/mid), type * sin(Pi/mid));
        for (int j = 0;j < lim; j += (mid << 1)) {
            Complex t(1, 0);
            for (int k = 0;k < mid; k++, t = t * T) {
                Complex x = a[j + k], y = t * a[mid + j + k];
                a[k + j] = x + y;
                a[mid + j + k] = x - y;
            }
        }
    }
}

int main() {
    read(n), read(m);
    for (int i = 0;i <= n; i++) {
        int x; read(x);
        A[i].x = x;
    }
    for (int i = 0;i <= m; i++) {
        int x; read(x);
        B[i].x = x;
    }
    while (lim <= (n + m)) lim <<= 1, L++;
    for (int i = 0;i < lim; i++)
    r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    
    FFT(A, 1); FFT(B, 1);
    for (int i = 0;i <= lim; i++) A[i] = A[i] * B[i];
    FFT(A, -1);
    for (int i = 0;i <= n + m; i++)
    printf ("%d ", (int)(A[i].x / lim + 0.5));
    return 0;
}

2.NTT快速数论变换

直接上代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int N = 3e6+6;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
inline int read(void) {
    int x = 0, f = 1;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+c-'0';
    return x * f;
}
int n, m, r[N], L;
ll A[N], B[N];
int lim = 1;
ll pow(ll di,ll mi) {
    ll ans = 1, a = di;
    while (mi) {
        if (mi & 1) ans = ans * a % P;
        a = a * a % P;
        mi >>= 1;
    }
    return ans;
}
void NTT(ll *a, int tag) {
    for (int i = 0;i < lim; i++) 
    if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim ;mid <<= 1) {
        ll wn = pow(tag == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int j = 0;j < lim; j += (mid << 1)) {
            ll w = 1;
            for (int k = 0;k < mid ; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = w * a[j+k+mid] % P;
                a[j+k] = (x + y) % P;
                a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
}
int main() {
    n = read(), m = read();
    for (int i = 0;i <= n; i++) A[i] = read();
    for (int j = 0;j <= m; j++) B[j] = read();
    while (lim <= n + m) L++, lim <<= 1;
    for (int i = 0;i < lim; i++)
    r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    NTT(A, 1), NTT(B, 1);
    for (int i = 0;i <= lim; i++) A[i] = (A[i] * B[i]) % P;
    NTT(A, -1);
    ll inv = pow(lim, P-2);
    for (int i = 0;i <= n + m; i++) 
    printf ("%d ", A[i] * inv % P);
    return 0;
}

3.多项式求逆

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
const int N = 3e5+5;
template <typename T>
void read(T &x) {
    x = 0; int f = 1;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1) + c-'0';
    x *= f;
}
ll A[N], n, lim = 1, L;
int r[N];
ll pow(ll di,ll mi) {
    ll ans = 1;
    while (mi) {
        if (mi & 1) ans = (ans * di) % P;
        di = (di * di) % P;
        mi >>= 1;
    }
    return ans;
}
void NTT(ll *a, int type) {
    for (int i = 0;i < lim; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = pow(type == 1 ? G : Gi, (P-1)/(mid<<1));
        for (int j = 0;j < lim;j += (mid << 1)) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = w * a[j+k+mid] % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = pow(lim, P-2);
    for (int i = 0;i < lim; i++)
    a[i] = a[i] * inv % P;
}
    
ll B[N], C[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = pow(a[0], P-2);
        return;
    }
    work((deg+1)>>1, a, b);
    lim = 1, L = 0;
    while (lim < (deg<<1)) lim <<= 1, L++;
    for (int i = 0;i < lim; i++) 
    r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) C[i] = a[i];
    for (int i = deg;i < lim; i++) C[i] = 0;
    NTT(C, 1), NTT(b, 1);
    for (int i = 0;i < lim; i++)
    b[i] = ((ll)2 - C[i] * b[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
int main() {
    read(n);
    for (int i = 0;i < n; i++) read(A[i]);
    work(n, A, B);
    for (int i = 0;i < n; i++) printf ("%lld ", B[i]);
    return 0;
}

4.分治FFT


#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi =  (P+1) / G;
const int N = 405000;
ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}


int lim, L;
ll r[N];
void NTT(ll *a,int type) {
    for (int i = 0;i < lim; i++) 
        if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid<<1));
        for (int i = 0;i < lim; i += mid << 1) {
            ll w = 1;
            for (int j = 0;j < mid; j++, w = w * wn % P) {
                ll x = a[j + i], y = a[mid + i + j] * w % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], f[N], b[N], g[N];

void solve(int l,int R) {
    if (l == R) return;
    int mid = (l + R) >> 1;
    solve(l, mid); lim = 1, L = 0;
    int len = R - l + 1;
    for (int i = l;i <= mid; i++) a[i-l] = f[i];
    for (int i = 0;i < len; i++) b[i] = g[i];
    len += mid - l + 1;
    while (lim <= len) lim <<= 1, L++;
    for (int i = 0;i < lim; i++) 
        r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
    NTT(a, 1), NTT(b, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * b[i] % P;
    NTT(a, -1);
    for (int i = mid + 1;i <= R; i++) f[i] = (f[i] + a[i-l]) % P;
    for (int i = 0;i < lim; i++) a[i] = b[i] = 0;
    solve(mid + 1, R);
}

int read(void) {
    int x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return x;
}

int n;
int main() {
    n = read();
    for (int i = 1;i < n; i++) g[i] = read();
    f[0] = 1;
    solve(0, n - 1);
    for (int i = 0;i < n; i++) printf ("%lld ", f[i]);
    return 0;
}

5.多项式ln：

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;
const int N = 505000;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1) / G;
int read(void) {
    int x = 0; bool f = 0;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return f ? -x : x;
}

ll fpw(ll di,ll mi) {
    ll ans = 1;
    while (mi) {
        if (mi & 1) ans = ans * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return ans;
}

ll A[N], B[N], C[N];
int r[N], n, lim, L;

void NTT(ll *a, int type) {
    for (int i = 0;i < lim; i++) 
        if (i < r[i]) swap(a[i], a[r[i]]);
    for (int j = 1;j < lim; j <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (j<<1));
        for (int k = 0;k < lim; k += j << 1) {
            ll w = 1;
            for (int p = 0;p < j; p++, w = w * wn % P) {
                ll x = a[p+k], y = a[p+k+j] * w % P;
                a[p+k] = (x + y) % P;
                a[p+k+j] = (x - y + P) % P;
            }
        }
    }
    if (~type) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) 
        a[i] = a[i] * inv % P;
}

void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = fpw(a[0], P-2);
        return;
    }
    work((deg+1)>>1, a, b);
    lim = 1, L = 0;
    while (lim < (deg<<1)) lim <<= 1, L++;
    for (int i = 0;i < lim; i++) 
        r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
    for (int i = 0;i < deg; i++) C[i] = a[i];
    for (int i = deg;i < lim; i++) C[i] = 0;
    NTT(C, 1); NTT(b, 1);
    for (int i = 0;i < lim; i++) 
    b[i] = ((ll)2 - C[i] * b[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}

int m;
void qiudoor(void) {
    for (int i = 1;i < n; i++) A[i-1] = A[i] * i % P;
    A[n-1] = 0;
}

void jinitaimei(void) {
    for (int i = n-1;i >= 1; i--) A[i] = A[i-1] * fpw(i, P-2) % P;
    A[0] = 0;
}

int main() {
    n  = read();
    for (int i = 0;i < n; i++) A[i] = read();
    work(n, A, B); qiudoor();
    lim = 1, L = 0; m = n;
    while (lim <= n + m) lim <<= 1, L++;
    NTT(A, 1); NTT(B, 1);
    for (int i = 0;i < lim; i++) A[i] = A[i] * B[i] % P;
    NTT(A, -1); 
    jinitaimei();
    for (int i = 0;i < n; i++) printf ("%lld ", A[i] % P);
    return 0;
}

6.多项式exp

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int N = 1005000;

int read(void) {
    int x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

int lim, L;
int r[N];
void NTT(ll *a,int type) {
    for (int i = 0;i < lim; i++)
        if (r[i] > i) swap(a[r[i]], a[i]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int i = 0;i < lim; i += mid << 1) {
            ll w = 1;
            for (int j = 0;j < mid; j++, w = w * wn % P) {
                int x = a[i + j], y = a[i + j + mid] * w % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], b[N], c[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = fpw(a[0], P-2);
        return;
    }
    work((deg+1) >> 1, a, b);
    lim = 1, L = 0;
    while (lim < (deg << 1)) lim <<= 1, L++;
    for (int i = 0;i < lim ;i++) 
        r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) c[i] = a[i];
    for (int i = deg; i < lim; i++) c[i] = 0;
    NTT(b, 1); NTT(c, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = (2 - b[i] * c[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
int n;
void qiudoor(ll *a, int deg) {
    for (int i = 1;i < n; i++) a[i-1] = a[i] * i % P;
    a[n-1] = 0;
}

ll inv[N];

void init(void) {
    inv[1] = 1;
    for (int i = 2;i <= n * 2; i++)
        inv[i] = (P - P / i * inv[P%i] % P) % P;
}

void jinitaimei(ll *a, int deg) {
    for (int i = n-1;i >= 1; i--) a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

ll A[N], B[N];

void get_ln(ll *a, ll n) {
    for (int i = 0;i < n*2; i++) B[i] = 0;
//  put(A, n), put(B, n*2);
    work(n, a, B); 
    qiudoor(a, n);
    NTT(a, 1); NTT(B, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * B[i] % P;
    NTT(a, -1); jinitaimei(a, n);
}

void solve(int deg, ll *a, ll *b) {
    if (deg == 1) return (void) (b[0] = 1);
    solve(deg>>1, a,  b);
    for (int i = 0;i < deg; i++) A[i] = b[i];
    get_ln(A, deg); 
    A[0] = (a[0] + 1 - A[0] + P) % P;
    for (int i = 1;i < deg; i++)
        A[i] = (a[i] - A[i] + P) % P;
    NTT(A, 1); NTT(b, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = b[i] * A[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}


ll f[N], g[N];
int main() {
    n = read(); init();
    for (int i = 0;i < n; i++) f[i] = read();
    lim = 1;
    while (lim <= n) lim <<= 1;
    solve(lim, f, g);
    for (int i = 0;i < n; i++) 
        printf ("%d ", g[i]);
    return 0;
}

7.多项式开根


#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int N = 1005000;

int read(void) {
    int x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

int lim, L;
int r[N];
void NTT(ll *a,int type) {
    for (int i = 0;i < lim; i++)
        if (r[i] > i) swap(a[r[i]], a[i]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int i = 0;i < lim; i += mid << 1) {
            ll w = 1;
            for (int j = 0;j < mid; j++, w = w * wn % P) {
                int x = a[i + j], y = a[i + j + mid] * w % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], b[N], c[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = fpw(a[0], P-2);
        return;
    }
    work((deg+1) >> 1, a, b);
    lim = 1, L = 0;
    while (lim < (deg << 1)) lim <<= 1, L++;
    for (int i = 0;i < lim ;i++) 
        r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) c[i] = a[i];
    for (int i = deg; i < lim; i++) c[i] = 0;
    NTT(b, 1); NTT(c, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = (2 - b[i] * c[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
int n;
void qiudoor(ll *a, int deg) {
    for (int i = 1;i < n; i++) a[i-1] = a[i] * i % P;
    a[n-1] = 0;
}

ll inv[N];

void init(void) {
    inv[1] = 1;
    for (int i = 2;i <= n * 2; i++)
        inv[i] = (P - P / i * inv[P%i] % P) % P;
}

void jinitaimei(ll *a, int deg) {
    for (int i = n-1;i >= 1; i--) a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

ll A[N], B[N];

void get_ln(ll *a, ll n) {
    for (int i = 0;i < n*2; i++) B[i] = 0;
//  put(A, n), put(B, n*2);
    work(n, a, B); 
    qiudoor(a, n);
    NTT(a, 1); NTT(B, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * B[i] % P;
    NTT(a, -1); jinitaimei(a, n);
}

void solve(int deg, ll *a, ll *b) {
    if (deg == 1) return (void) (b[0] = 1);
    solve(deg>>1, a,  b);
    lim = 1, L = 0;
    for (int i = 0;i < deg; i++) A[i] = b[i];
    get_ln(A, deg); 
    A[0] = (a[0] + 1 - A[0] + P) % P;
    for (int i = 1;i < deg; i++)
        A[i] = (a[i] - A[i] + P) % P;
    NTT(A, 1); NTT(b, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = b[i] * A[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}


ll f[N], g[N];
int main() {
    n = read(); init();
    for (int i = 0;i < n; i++) f[i] = read();
    get_ln(f, n);
    for (int i = 0;i < n; i++) f[i] = f[i] * inv[2] % P;
    lim = 1;
    while (lim <= n) lim <<= 1;
    solve(lim, f, g);
    for (int i = 0;i < n; i++) 
        printf ("%d ", g[i]);
    return 0;
}

8.多项式除法


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3;
const int Gi = (P+1)/G;
const int N = 3e5+5;
template <typename T>
void read(T &x) {
    x = 0; int f = 1;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1) + c-'0';
    x *= f;
}
ll A[N], n, lim = 1, L;
int r[N];
ll pow(ll di,ll mi) {
    ll ans = 1;
    while (mi) {
        if (mi & 1) ans = (ans * di) % P;
        di = (di * di) % P;
        mi >>= 1;
    }
    return ans;
}
void NTT(ll *a, int type) {
    for (int i = 0;i < lim; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = pow(type == 1 ? G : Gi, (P-1)/(mid<<1));
        for (int j = 0;j < lim;j += (mid << 1)) {
            ll w = 1;
            for (int k = 0;k < mid; k++, w = (w * wn) % P) {
                ll x = a[j+k], y = w * a[j+k+mid] % P;
                a[j+k] = (x + y) % P, a[j+k+mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = pow(lim, P-2);
    for (int i = 0;i < lim; i++)
    a[i] = a[i] * inv % P;
}

ll B[N], C[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = pow(a[0], P-2);
        return;
    }
    work((deg+1)>>1, a, b);
    lim = 1, L = 0;
    while (lim < (deg<<1)) lim <<= 1, L++;
    for (int i = 0;i < lim; i++) 
    r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) C[i] = a[i];
    for (int i = deg;i < lim; i++) C[i] = 0;
    NTT(C, 1), NTT(b, 1);
    for (int i = 0;i < lim; i++)
    b[i] = ((ll)2 - C[i] * b[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
ll f[N], g[N], q[N], ff[N], gg[N];
int m;
int main() {
    freopen("hs.in","r",stdin);
    freopen("hs.out","w",stdout);
    read(n), read(m);
    for (int i = n;i >= 0; i--) read(f[i]), ff[n-i] = f[i];
    for (int j = m;j >= 0; j--) read(g[j]), gg[m-j] = g[j];
    for (int i = n - m + 1;i <= n; i++) f[i] = g[i] = 0;
    work(n-m+1, g, B);
    NTT(B, 1); NTT(f, 1);
    for (int i = 0;i < lim; i++) q[i] = f[i] * B[i] % P;
    NTT(q, -1);
    reverse(q, q + n - m + 1);
    for (int i = 0;i <= n - m; i++) printf ("%lld ", q[i]);
    lim = 1, L = 0;
    while (lim <= n) lim <<= 1, L++;
    for (int i = n - m + 1;i < lim; i++) q[i] = 0;
    for (int i = 0;i < lim; i++) r[i] = (r[i>>1]>>1) | ((i&1)<<(L-1));
    NTT(gg, 1); NTT(q, 1);
    for (int i = 0;i < lim; i++) gg[i] = gg[i] * q[i] % P;
    NTT(gg, -1);
    putchar('\n');
    for (int i = 0;i < m; i++) printf ("%lld ", (ff[i] - gg[i] + P) % P);
    return 0;
}

9.拉格朗日插值

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int P = 998244353;
const int N = 3e5+5;
template <typename T>
void read(T &x) {
    x = 0; int f = 1;
    char c = getchar();
    for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1) + c-'0';
    x *= f;
}

ll fpw(ll x,ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * x % P;
        x = x * x % P;
        mi >>= 1;
    }
    return res;
}

int x[N], y[N], n, k;
int main() {
    read(n), read(k);
    for (int i = 1;i <= n; i++) read(x[i]), read(y[i]);
    ll res = 0;
    for (int i = 1;i <= n; i++) {
        ll ans1 = 1, ans2 = 1;
        for (int j = 1;j <= n; j++) {
            if (j == i) continue;
            ans1 = ans1 * (k - x[j]) % P;
            ans2 = ans2 * (x[i] - x[j]) % P;
        }
        res += y[i] * ans1 % P * fpw(ans2, P-2) % P;
    }
    res %= P;
    if (res < 0) res += P;
    cout << res << endl;
    return 0;
}

10. 多项式快速幂

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const int P = 998244353;
const int G = 3, Gi = (P+1) / G;
const int N = 1005000;

int read(void) {
    int x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^48);
    return x;
}

ll fpw(ll di, ll mi) {
    ll res = 1;
    while (mi) {
        if (mi & 1) res = res * di % P;
        di = di * di % P;
        mi >>= 1;
    }
    return res;
}

int lim, L;
int r[N];
void NTT(ll *a,int type) {
    for (int i = 0;i < lim; i++)
        if (r[i] > i) swap(a[r[i]], a[i]);
    for (int mid = 1;mid < lim; mid <<= 1) {
        ll wn = fpw(type == 1 ? G : Gi, (P-1) / (mid << 1));
        for (int i = 0;i < lim; i += mid << 1) {
            ll w = 1;
            for (int j = 0;j < mid; j++, w = w * wn % P) {
                int x = a[i + j], y = a[i + j + mid] * w % P;
                a[i + j] = (x + y) % P;
                a[i + j + mid] = (x - y + P) % P;
            }
        }
    }
    if (type == 1) return;
    ll inv = fpw(lim, P-2);
    for (int i = 0;i < lim; i++) a[i] = a[i] * inv % P;
}

ll a[N], b[N], c[N];
void work(int deg, ll *a, ll *b) {
    if (deg == 1) {
        b[0] = fpw(a[0], P-2);
        return;
    }
    work((deg+1) >> 1, a, b);
    lim = 1, L = 0;
    while (lim < (deg << 1)) lim <<= 1, L++;
    for (int i = 0;i < lim ;i++) 
        r[i] = (r[i>>1]>>1) | ((i&1) << (L-1));
    for (int i = 0;i < deg; i++) c[i] = a[i];
    for (int i = deg; i < lim; i++) c[i] = 0;
    NTT(b, 1); NTT(c, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = (2 - b[i] * c[i] % P + P) % P * b[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}
int n;
void qiudoor(ll *a, int deg) {
    for (int i = 1;i < n; i++) a[i-1] = a[i] * i % P;
    a[n-1] = 0;
}

ll inv[N];

void init(void) {
    inv[1] = 1;
    for (int i = 2;i <= n * 2; i++)
        inv[i] = (P - P / i * inv[P%i] % P) % P;
}

void jinitaimei(ll *a, int deg) {
    for (int i = n-1;i >= 1; i--) a[i] = a[i-1] * inv[i] % P;
    a[0] = 0;
}

ll A[N], B[N];

void get_ln(ll *a, ll n) {
    for (int i = 0;i < n*2; i++) B[i] = 0;
//  put(A, n), put(B, n*2);
    work(n, a, B); 
    qiudoor(a, n);
    NTT(a, 1); NTT(B, 1);
    for (int i = 0;i < lim; i++) a[i] = a[i] * B[i] % P;
    NTT(a, -1); jinitaimei(a, n);
}

void solve(int deg, ll *a, ll *b) {
    if (deg == 1) return (void) (b[0] = 1);
    solve(deg>>1, a,  b);
    for (int i = 0;i < deg; i++) A[i] = b[i];
    get_ln(A, deg); 
    A[0] = (a[0] + 1 - A[0] + P) % P;
    for (int i = 1;i < deg; i++)
        A[i] = (a[i] - A[i] + P) % P;
    NTT(A, 1); NTT(b, 1);
    for (int i = 0;i < lim; i++) 
        b[i] = b[i] * A[i] % P;
    NTT(b, -1);
    for (int i = deg;i < lim; i++) b[i] = 0;
}

ll f[N], g[N];

ll get_k(void) {
    ll x = 0; char c = getchar();
    while (!isdigit(c)) c=getchar();
    for (;isdigit(c);c=getchar()) {
        x=(x<<3)+(x<<1)+(c^48);
        if (x >= P) x %= P;
    }
    return x;
}
ll k;
int main() {
    n = read(); init(); k = get_k();
    for (int i = 0;i < n; i++) f[i] = read();
    get_ln(f, n);
    for (int i = 0;i < n; i++) f[i] = f[i] * k % P;
    lim = 1;
    while (lim <= n) lim <<= 1;
    solve(lim, f, g);
    for (int i = 0;i < n; i++) 
        printf ("%d ", g[i]);
    return 0;
}

多项式全家桶

标签：朴素   git   dft   离散傅里叶变换   lin   快速   傅里叶变换   cpp   lex   

原文地址：https://www.cnblogs.com/Hs-black/p/12004617.html
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