"Basis polynomial algorithm summary"

Others 2019-08-28 06:14:34 views: null

<> Yes

<First update> At the request of the coach began to learn polynomial algorithm, but because they can not be integral and derivative first iteration of Newton polynomial, exponential polynomial, polynomial power function, power polynomial quickly and so goo out, so this is a blog that summarizes the content of other polynomial basis.

<Text>

Fast Fourier Transform

\ (An FFT \) , fast Fourier transform, may be \ (O (n \ log_2n) \) within a time calculating polynomial multiplication.

First recall \ (FFT \) idea, first coefficients of the polynomial expressions, if the direct calculation, time complexity is \ (O (n ^ 2) \) can be optimized using divide and conquer algorithm to \ (O (n-log_23 ^ {}) \) , typically no other better the algorithm.

And another is to consider the expression, polynomial expressions point value, this can be achieved \ (O (n) \) multiplication.

However, to achieve expression of the coefficient conversion method and point value expressions are evaluated using conventional methods and the Lagrange polynomial \ (O (n ^ 2) \) , and meaningless.

Later, a divide and conquer algorithm appeared, we can fit into the evaluation when \ (n \) unit root, depending on the nature of unit root, we can calculate only half the worth to the other half, so recursion partition resolve to achieve time complexity \ (O (n-\ log_2n) \) .

When it comes to unit root, we must mention the basics of the complex, you can look at this blog .

Then that is the nature of the unit root:

\(1.\) \(\omega_{n}^k=\cos k\frac{2\pi}{n}+i\sin k\frac{2\pi}{n}\)

\(2.\) \(\omega_{2n}^{2k}=\omega_n^k\)

\(3.\) \(\omega_{n}^{k+\frac{n}{2}}=-\omega_{n}^{k}\)

\(4.\) \(k\not=0\)时，\(\sum_{i=0}^{n-1}(\omega_{n}^{k})^i=0\)，\(k=0\)时，\(\sum_{i=0}^{n-1}(\omega_{n}^{k})^i=n\)。

Then the polynomial parity index by classification, you can partition a.

By constructing understood, we can also be converted in the same manner as back point value expressions coefficient expressions, it is divided by the size of a sequence length.

Finally, the last is to recursively \ (FFT \) with iterative implementation is a little small binary \ (Trick \) , called the butterfly theorem, to find the law to do.

You can see this blog .

\(Code:\)

#include <bits/stdc++.h>
using namespace std;
const int N = (1<<22) + 20;
namespace Mathcalc
{
    const double PI = acos(-1.0) , eps = 5e-3;
    struct Complex
    {
        double x,y;
        Complex ( double _x = 0 , double _y = 0 ) { x = _x , y = _y; }
        friend Complex operator + (Complex a,Complex b) { return Complex( a.x + b.x , a.y + b.y ); }
        friend Complex operator - (Complex a,Complex b) { return Complex( a.x - b.x , a.y - b.y ); }
        friend Complex operator * (Complex a,Complex b) { return Complex( a.x*b.x - a.y*b.y , a.x*b.y + a.y*b.x ); }
    };
}
using namespace Mathcalc;
inline int read(void)
{
    int x = 0 , w = 0; char ch = ' ';
    while ( !isdigit(ch) ) w |= ch=='-' , ch = getchar();
    while ( isdigit(ch) ) x = x*10 + ch-48 , ch = getchar();
    return w ? -x : x;
}
int n,m,lenth,cnt,rev[N];
Complex a[N],b[N],c[N];
inline void input(void)
{
    n = read() , m = read();
    for (int i=0;i<=n;i++) a[i].x = read();
    for (int i=0;i<=m;i++) b[i].x = read();
}
inline void calcrev(void)
{
    lenth = 1;
    while ( lenth <= n + m ) lenth <<= 1 , cnt++;
    for (int i=0;i<lenth;i++)
        rev[i] = ( rev[i>>1] >> 1 ) | ( (i&1) << (cnt-1) );
}
inline void FFT(Complex *p,int op)
{
    for (int i=0;i<lenth;i++)
        if ( i < rev[i] ) swap( p[i] , p[rev[i]] );
    for (int i=1;i<lenth;i<<=1)
    {
        Complex w ( cos(PI/i) , op * sin(PI/i) );
        for (int j=0;j<lenth;j+=i<<1)
        {
            Complex omega ( 1 , 0 );
            for (int k=0;k<i;k++)
            {
                Complex x = p[j+k] , y = omega * p[i+j+k];
                p[j+k] = x + y , p[i+j+k] = x - y;
                omega = omega * w;
            }
        }
    }
    if ( op == -1 )
        for (int i=0;i<lenth;i++)
            p[i].x /= lenth;
}
int main(void)
{
    input();
    calcrev();
    FFT( a , 1 ) , FFT( b , 1 );
    for (int i=0;i<lenth;i++)
        c[i] = a[i] * b[i];
    FFT( c , -1 );
    for (int i=0;i<=n+m;i++)
        printf("%d ",int(c[i].x + eps));
    return 0;
}

Number Theoretic Transform

\ (NTT \) , fast number theory transformation, may be (O (n \ log_2n) \ ) \ calculated polynomial multiplying the significance of mold time.

思路和\(FFT\)是一模一样的，关键在于我们在模意义下找到了单位根的替代品：原根，满足单位根的所有性质。

不需要用到太多数论知识，可以看这篇博客和这篇博客。

然后按照\(FFT\)的方式来就可以了。

但是\(NTT\)的模数是有限定的，可以看这个。

\(Code:\)

#include <bits/stdc++.h>
using namespace std;
const int N = (1<<22)+20;
namespace Mathcalc
{
    const int Mod = 998244353 , G = 3 , INVG = 332748118;
    inline int add(int a,int b) { return a + b >= Mod ? a + b - Mod : a + b; }
    inline int sub(int a,int b) { return a - b < 0 ? a - b + Mod : a - b; }
    inline int mul(int a,int b) { return 1LL * a * b % Mod; }
    inline void Mul(int &a,int b) { a = mul( a , b ); }
    inline int quickpow(int a,int b) { int res = 1; for (;b;Mul(a,a),b>>=1) if ( b & 1 ) Mul( res , a ); return res; }
    inline int inv(int a) { return quickpow( a , Mod-2 ); }
};
using namespace Mathcalc;
inline int read(void)
{
    int x = 0 , w = 0; char ch = ' ';
    while ( !isdigit(ch) ) w |= ch=='-' , ch = getchar();
    while ( isdigit(ch) ) x = x*10 + ch-48 , ch = getchar();
    return w ? -x : x;
}
int n,m,lenth,cnt,rev[N];
int a[N],b[N],c[N];
inline void input(void)
{
    n = read() - 1 , m = read() - 1;
    for (int i=0;i<=n;i++) a[i] = read();
    for (int i=0;i<=m;i++) b[i] = read();
}
inline void calcrev(void)
{
    lenth = 1;
    while ( lenth <= n + m ) lenth <<= 1 , cnt++;
    for (int i=0;i<lenth;i++)
        rev[i] = ( rev[i>>1] >> 1 ) | ( (i&1) << (cnt-1) );
}
inline void NTT(int *p,int op)
{
    for (int i=0;i<lenth;i++)
        if ( i < rev[i] ) swap( p[i] , p[rev[i]] );
    for (int i=1;i<lenth;i<<=1)
    {
        int w = quickpow( op == 1 ? G : INVG , (Mod-1) / (i<<1) );
        for (int j=0;j<lenth;j+=i<<1)
        {
            int omega = 1;
            for (int k=0;k<i;k++)
            {
                int x = p[j+k] , y = mul( omega , p[i+j+k] );
                p[j+k] = add( x , y ) , p[i+j+k] = sub( x , y );
                Mul( omega , w );
            }
        }
    }
    if ( op == -1 )
    {
        int val = inv( lenth );
        for (int i=0;i<lenth;i++)
            Mul( p[i] , val );
    }
}
int main(void)
{
    input();
    calcrev();
    NTT( a , 1 ) , NTT( b , 1 );
    for (int i=0;i<lenth;i++)
        c[i] = mul( a[i] , b[i] );
    NTT( c , -1 );
    for (int i=0;i<=n+m;i++)
        printf("%d ",c[i]);
    return 0;
}

分治FFT

我们知道，\(FFT/NTT\)是可以求形容\(c_k=\sum_{i+j=k}a_i\times b_j\)的卷积的，但是如果是长这样的卷积呢？

\[f_k=\sum_{i+j=k}f_i\times g_j\]

仿佛套一个\(cdq\)分治算贡献就可以了。

没错，事实上也就是这样，中间的卷积还是要\(NTT\)，时间复杂度\(O(n\log^2n)\)。

\(Cocd:\)

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+20 , SIZE = (1<<18)+20;
const int Mod = 998244353 , G = 3 , INVG = 332748118;
int n,cnt,lenth,rev[SIZE];
long long f[N],g[N],a[SIZE],b[SIZE],c[SIZE];
inline int read(void)
{
    int x = 0 , w = 0; char ch = ' ';
    while ( !isdigit(ch) ) w |= ch=='-' , ch = getchar();
    while ( isdigit(ch) ) x = x*10 + ch-48 , ch = getchar();
    return w ? -x : x;
}
inline void input(void)
{
    n = read() - 1 , f[0] = 1;
    for (int i=1;i<=n;i++)
        g[i] = read();
}
inline long long quickpow(long long a,long long b)
{
    long long res = 1;
    while ( b )
    {
        if ( 1 & b ) res = res * a % Mod;
        b >>= 1 , a = a * a % Mod;
    }
    return res;
}
inline void NTT(long long *p,int op)
{
    for (int i=0;i<lenth;i++)
        if ( i < rev[i] ) swap( p[i] , p[rev[i]] );
    for (int i=1;i<lenth;i<<=1)
    {
        long long w = quickpow( op == 1 ? G : INVG , (Mod-1) / (i<<1) );
        for (int j=0;j<lenth;j+=i<<1)
        {
            long long omega = 1;
            for (int k=0;k<i;k++)
            {
                long long x = p[j+k] , y = omega * p[i+j+k] % Mod;
                p[j+k] = ( x + y ) % Mod;
                p[i+j+k] = ( ( x - y ) % Mod + Mod ) % Mod;
                omega = omega * w % Mod;
            }
        }
    }
}
inline void cdq(int l,int r)
{
    if ( l == r ) return void();
    int mid = l+r >> 1 , len = (r-l-1) + (mid-l);
    cdq( l , mid );
    lenth = 1 , cnt = 0;
    while ( lenth <= len ) lenth <<= 1 , cnt++;
    for (int i=0;i<lenth;i++)
        rev[i] = ( rev[i>>1] >> 1 ) | ( (i&1) << (cnt-1) ),
        a[i] = b[i] = c[i] = 0;
    for (int i=l;i<=mid;i++) a[i-l] = f[i];
    for (int i=1;i<=r-l;i++) b[i-1] = g[i];
    NTT( a , 1 ) , NTT( b , 1 );
    for (int i=0;i<lenth;i++) c[i] = a[i] * b[i] % Mod;
    NTT( c , -1 );
    long long inv = quickpow( lenth , Mod-2 );
    for (int i=0;i<=len;i++)
        c[i] = c[i] * inv % Mod;
    for (int i=mid+1;i<=r;i++)
        f[i] = ( f[i] + c[i-l-1] ) % Mod;
    cdq( mid + 1 , r );
}
int main(void)
{
    input();
    cdq( 0 , n );
    for (int i=0;i<=n;i++)
        printf("%lld ",f[i]);
    puts("");
    return 0;
}

Fast Walsh-Hadamard Transform

\(FWT\)，快速沃尔什变换，用于计算位运算卷积。

形式是这样的\(c_k=\sum_{i\ op\ j=k}a_i\times b_j\)，\(op\)是\(and,or,xor\)中的一个位运算。

思路和\(FWT\)是一样的，考虑先对系数表达法做一点变换，然后点乘\(O(n)\)解决问题，再逆变换回来就可以了。

好像不要那么多公式？看这三篇博客吧：1，2，3。

\(Code:\)

#include <bits/stdc++.h>
using namespace std;
const int N = 20;
namespace Mathcalc
{
    const int Mod = 998244353 , INV2 = 499122177;
    inline int add(int a,int b) { return a + b >= Mod ? a + b - Mod : a + b; }
    inline int sub(int a,int b) { return a - b < 0 ? a - b + Mod : a - b; }
    inline int mul(int a,int b) { return 1LL * a * b % Mod; }
    inline void Add(int &a,int b) { a = add( a , b ); }
    inline void Sub(int &a,int b) { a = sub( a , b ); }
    inline void Mul(int &a,int b) { a = mul( a , b ); }
};
using namespace Mathcalc;
int n; int a[1<<N],b[1<<N],c[1<<N];
inline void input(void)
{
    scanf("%d",&n);
    for (int i=0;i<1<<n;i++)
        scanf("%d",&a[i]);
    for (int i=0;i<1<<n;i++)
        scanf("%d",&b[i]);
}
inline void FWTor(int *p,int op)
{
    for (int i=1;i<1<<n;i<<=1)
        for (int j=0;j<1<<n;j+=i<<1)
            for (int k=0;k<i;k++)
                if ( op == 1 ) Add( p[i+j+k] , p[j+k] );
                else Sub( p[i+j+k] , p[j+k] );
}
inline void FWTand(int *p,int op)
{
    for (int i=1;i<1<<n;i<<=1)
        for (int j=0;j<1<<n;j+=i<<1)
            for (int k=0;k<i;k++)
                if ( op == 1 ) Add( p[j+k] , p[i+j+k] );
                else Sub( p[j+k] , p[i+j+k] );
}
inline void FWTxor(int *p,int op)
{
    for (int i=1;i<1<<n;i<<=1)
        for (int j=0;j<1<<n;j+=i<<1)
            for (int k=0;k<i;k++)
            {
                int x = p[j+k] , y = p[i+j+k];
                p[j+k] = mul( op , add( x , y ) );
                p[i+j+k] = mul( op , sub( x , y ) );
            }
}
inline void solve(void)
{
    FWTor( a , 1 ) , FWTor( b , 1 );
    for (int i=0;i<1<<n;i++) c[i] = mul( a[i] , b[i] );
    FWTor( c , -1 );
    for (int i=0;i<1<<n;i++) printf("%d ",c[i]);
    puts("") , FWTor( a , -1 ) , FWTor( b , -1 );

    FWTand( a , 1 ) , FWTand( b , 1 );
    for (int i=0;i<1<<n;i++) c[i] = mul( a[i] , b[i] );
    FWTand( c , -1 );
    for (int i=0;i<1<<n;i++) printf("%d ",c[i]);
    puts("") , FWTand( a , -1 ) , FWTand( b , -1 );

    FWTxor( a , 1 ) , FWTxor( b , 1 );
    for (int i=0;i<1<<n;i++) c[i] = mul( a[i] , b[i] );
    FWTxor( c , INV2 );
    for (int i=0;i<1<<n;i++) printf("%d ",c[i]);
    puts("") , FWTxor( a , INV2 ) , FWTxor( b , INV2 );
}
int main(void)
{
    input();
    solve();
    return 0;
}

多项式求逆

给定多项式\(F(x)\)，求多项式\(G(x)\)，满足\(F(x)G(x)\equiv1\pmod {x^n}\)。

当多项式只有一个常数项的时候，求逆元即可。

反之，考虑一个倍增算法：

假设\(F\times G'\equiv 1\pmod{x^{\frac{n}{2}}}\)，我们要求\(F\times G\equiv1\pmod{x^n}\)。

\[F\times(G-G')\equiv 0\pmod {x^{\frac{n}{2}}}\\ \ \\G-G'\equiv 0\pmod {x^{\frac{n}{2}}}\\\ \\(G-G')^2\equiv 0\pmod {x^n}\\\ \\G^2-2GG'+G'^2\equiv 0\pmod{x^n}\\ \ \\F\times(G^2-2GG'+G'^2)\equiv 0\pmod{x^n}\\\ \\G-2G'+FG'^2\equiv 0\pmod{x^n}\\ \ \\G\equiv2G'+FG'^2\pmod{x^n}\]

结合\(NTT\)，就可以倍增计算了，时间复杂度\(O(n\log_2n)\)。

\(Code:\)

#include <bits/stdc++.h>
using namespace std;
const int N = (1<<20)+20;
namespace Mathcalc
{
    const int Mod = 998244353 , G = 3 , INVG = 332748118;
    inline int add(int a,int b) { return a + b >= Mod ? a + b - Mod : a + b; }
    inline int sub(int a,int b) { return a - b < 0 ? a - b + Mod : a - b; }
    inline int mul(int a,int b) { return 1LL * a * b % Mod; }
    inline void Add(int &a,int b) { a = add( a , b ); }
    inline void Sub(int &a,int b) { a = sub( a , b ); }
    inline void Mul(int &a,int b) { a = mul( a , b ); }
    inline int quickpow(int a,int b) { int res = 1; for (;b;Mul(a,a),b>>=1) if ( b & 1 ) Mul( res , a ); return res; }
    inline int inv(int a) { return quickpow( a , Mod-2 ); }
};
using namespace Mathcalc;
inline int read(void)
{
    int x = 0 , w = 0; char ch = ' ';
    while ( !isdigit(ch) ) w |= ch=='-' , ch = getchar();
    while ( isdigit(ch) ) x = x*10 + ch-48 , ch = getchar();
    return w ? -x : x;
}
int temp[N],rev[N];
inline void calcrev(int len,int cnt)
{
    for (int i=0;i<len;i++)
        rev[i] = ( rev[i>>1] >> 1 ) | ( (i&1) << (cnt-1) );
}
inline void NTT(int *p,int op,int len)
{
    for (int i=0;i<len;i++)
        if ( i < rev[i] ) swap( p[i] , p[rev[i]] );
    for (int i=1;i<len;i<<=1)
    {
        int w = quickpow( op == 1 ? G : INVG , (Mod-1) / (i<<1) );
        for (int j=0;j<len;j+=i<<1)
        {
            int omega = 1;
            for (int k=0;k<i;k++)
            {
                int x = p[j+k] , y = mul( omega , p[i+j+k] );
                p[j+k] = add( x , y ) , p[i+j+k] = sub( x , y );
                Mul( omega , w );
            }
        }
    }
    if ( op == -1 )
    {
        int val = inv( len );
        for (int i=0;i<len;i++)
            Mul( p[i] , val );
    }
}
inline void Polyinv(int *a,int *b,int n)
{
    int p = 1; while ( p < n ) p <<= 1;
    for (int i=0;i<p;i++) b[i] = 0;
    b[0] = inv( a[0] );
    for (int len=1,cnt=0;len<=p;len<<=1,cnt++)
    {
        calcrev( len<<1 , cnt+1 );
        for (int i=0;i<len;i++) temp[i] = a[i];
        for (int i=len;i<len<<1;i++) temp[i] = 0;
        NTT( b , 1 , len<<1 ) , NTT( temp , 1 , len<<1 );
        for (int i=0;i<len<<1;i++)
            Mul( b[i] , sub( 2 , mul( temp[i] , b[i] ) ) );
        NTT( b , -1 , len<<1 );
        for (int i=len;i<len<<1;i++) b[i] = 0;
    }
    for (int i=n;i<p;i++) b[i] = 0;
}
int n,m,a[N],b[N];
inline void input(void)
{
    n = read() , m = read();
    for (int i=0;i<n;i++) a[i] = read();
}
int main(void)
{
    input();
    Polyinv( a , b , m );
    for (int i=0;i<m;i++)
        printf("%d ",b[i]);
    return 0;
}

多项式除法

给定\(n\)次多项式\(F(x)\)和\(m\)次多项式\(G(x)\)，求多项式\(Q(x),R(x)\)，满足：

\(1.\) \(Q(x)\)为\(n-m\)次多项式，\(R(x)\)的次数小于\(m\)。
\(2.\) \(F(x)=G(x)\times Q(x)+R(x)\)。

考虑一个操作\(rev\)，对于多项式\(F(x)\)，\(F^{rev}(x)=x^nF(\frac{1}{x})\)，就是系数翻转。

然后考虑推一波式子：
\[ F(x)=Q(x)\times G(x)+R(x) \\\ \\ F\left (\frac{1}{x} \right )=G\left (\frac{1}{x} \right ) \times Q\left (\frac{1}{x} \right )+R\left (\frac{1}{x} \right ) \\ \ \\ x^nF\left (\frac{1}{x} \right )=x^mG\left (\frac{1}{x} \right ) \times x^{n-m} Q\left (\frac{1}{x} \right )+x^{n-m+1}x^{m-1}R\left (\frac{1}{x} \right ) \\ \ \\ F^{rev}(x)=Q^{rev}(x)\times G^{rev}(x)+x^{n-m+1}R^{rev}(x) \\ \ \\ F^{rev}(x)=Q^{rev}(x)\times G^{rev}(x)\pmod{x^{n-m+1}} \\ \ \\ Q^{rev}(x)=F^{rev}(x)\times (G^{rev})^{-1}(x)\pmod{x^{n-m+1}} \]

考虑对\(G^{rev}\)做一个多项式求逆，然后再用\(NTT\)把它和\(F^{rev}\)乘起来就是\(Q^{rev}\)了。

然后\(R\)直接算就可以了：

\[R(x)=F(x)-Q(x)\times G(x)\]

\(Code:\)

#include <bits/stdc++.h>
using namespace std;
const int N = (1<<20)+20;
namespace Mathcalc
{
    const int Mod = 998244353 , G = 3 , INVG = 332748118;
    inline int add(int a,int b) { return a + b >= Mod ? a + b - Mod : a + b; }
    inline int sub(int a,int b) { return a - b < 0 ? a - b + Mod : a - b; }
    inline int mul(int a,int b) { return 1LL * a * b % Mod; }
    inline void Add(int &a,int b) { a = add( a , b ); }
    inline void Sub(int &a,int b) { a = sub( a , b ); }
    inline void Mul(int &a,int b) { a = mul( a , b ); }
    inline int quickpow(int a,int b) { int res = 1; for (;b;Mul(a,a),b>>=1) if ( b & 1 ) Mul( res , a ); return res; }
    inline int inv(int a) { return quickpow( a , Mod-2 ); }
};
using namespace Mathcalc;
inline int read(void)
{
    int x = 0 , w = 0; char ch = ' ';
    while ( !isdigit(ch) ) w |= ch=='-' , ch = getchar();
    while ( isdigit(ch) ) x = x*10 + ch-48 , ch = getchar();
    return w ? -x : x;
}
int temp[N],t1[N],t2[N],rev[N];
inline void calcrev(int len,int cnt)
{
    for (int i=0;i<len;i++)
        rev[i] = ( rev[i>>1] >> 1 ) | ( (i&1) << (cnt-1) );
}
inline void NTT(int *p,int op,int len)
{
    for (int i=0;i<len;i++)
        if ( i < rev[i] ) swap( p[i] , p[rev[i]] );
    for (int i=1;i<len;i<<=1)
    {
        int w = quickpow( op == 1 ? G : INVG , (Mod-1) / (i<<1) );
        for (int j=0;j<len;j+=i<<1)
        {
            int omega = 1;
            for (int k=0;k<i;k++)
            {
                int x = p[j+k] , y = mul( omega , p[i+j+k] );
                p[j+k] = add( x , y ) , p[i+j+k] = sub( x , y );
                Mul( omega , w );
            }
        }
    }
    if ( op == -1 )
    {
        int val = inv( len );
        for (int i=0;i<len;i++)
            Mul( p[i] , val );
    }
}
inline void Polyinv(int *a,int *b,int n)
{
    int p = 1; while ( p < n ) p <<= 1;
    for (int i=0;i<p;i++) b[i] = 0;
    b[0] = inv( a[0] );
    for (int len=1,cnt=0;len<=p;len<<=1,cnt++)
    {
        calcrev( len<<1 , cnt+1 );
        for (int i=0;i<len;i++) temp[i] = a[i];
        for (int i=len;i<len<<1;i++) temp[i] = 0;
        NTT( b , 1 , len<<1 ) , NTT( temp , 1 , len<<1 );
        for (int i=0;i<len<<1;i++)
            Mul( b[i] , sub( 2 , mul( temp[i] , b[i] ) ) );
        NTT( b , -1 , len<<1 );
        for (int i=len;i<len<<1;i++) b[i] = 0;
    }
    for (int i=n;i<p;i++) b[i] = 0;
}
inline void Polymul(int *a,int *b,int *c,int len)
{
    NTT( a , 1 , len ) , NTT( b , 1 , len );
    for (int i=0;i<len;i++)
        c[i] = mul( a[i] , b[i] );
    NTT( c , -1 , len );
}
inline void Polydiv(int *f,int *g,int *q,int *r,int n,int m)
{
    int len = 1 , cnt = 0;
    while ( len < 2*(n-m+1) ) len <<= 1 , cnt++;
    for (int i=0;i<n-m+1;i++) t2[i] = g[m-i-1];
    for (int i=n-m+1;i<len;i++) t2[i] = 0;
    Polyinv( t2 , t1 , n-m+1 );
    for (int i=0;i<n-m+1;i++) t2[i] = f[n-i-1];
    for (int i=n-m+1;i<len;i++) t2[i] = 0;
    calcrev( len , cnt ) , Polymul( t1 , t2 , q , len );
    reverse( q , q+n-m+1 );

    len = 1 , cnt = 0;
    while ( len < n ) len <<= 1 , cnt++;
    for (int i=0;i<n-m+1;i++) t1[i] = q[i];
    for (int i=n-m+1;i<len;i++) t1[i] = 0;
    for (int i=0;i<m;i++) t2[i] = g[i];
    for (int i=m;i<len;i++) t2[i] = 0;
    calcrev( len , cnt ) , Polymul( t1 , t2 , r , len );
    for (int i=0;i<len;i++)
        r[i] = sub( f[i] , r[i] );
}
int n,m,f[N],g[N],q[N],r[N];
int main(void)
{
    n = read() , m = read();
    for (int i=0;i<n;i++) f[i] = read();
    for (int i=0;i<m;i++) g[i] = read();
    Polydiv( f , g , q , r , n , m );
    for (int i=0;i<n-m+1;i++)
        printf("%d ",q[i]);
    puts("");
    for (int i=0;i<m-1;i++)
        printf("%d ",r[i]);
    puts("");
    return 0;
}

<后记>

Guess you like

Origin www.cnblogs.com/Parsnip/p/11421041.html
Recommended
The United States plans to restrict the export of large AI models to China and Russia
Apple to reach agreement with OpenAI to bring ChatGPT to iPhone
Ranking
whisper-webui installation tutorial is silky and easy to use
[Base] Laravel concepts laravel basis, the custom service provider: Contracts, ServiceContainer, ServiceProvider, Facades relations
Import torchvision error problem solving DLL: module not found
observer & watch & notify = pub & sub
A small turntable program [HTML + CSS + JS]
CorelDRAW 2018 shortcuts Daquan
Supervise el botón de menú para lograr un gatillo de presión prolongada
JS将时间秒转换成天小时分钟秒的字符串
RIP basic configuration
[Deleted] solution to a problem a few questions (Noip1994)
Daily
More
2024-05-11(32)
2024-05-10(34)
2024-05-09(32)
2024-05-08(18)
2024-05-07(34)
2024-05-06(6)
2024-05-05(0)
2024-05-04(18)
2024-05-03(8)
2024-05-02(0)

Code World

© 2018 codetd.com