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  • FFT学习

    看了一天的多项式的内容,看博客的时候好像还是那么回事,一看题,哇塞,发现我其实连卷积是啥都没看懂。

    qtdydb,背板子。 不知道卷积是啥就很伤了。

    P3803 【模板】多项式乘法(FFT)

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    typedef long long LL;
    
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
    
    const int N=4e6+5;
    const double pi=acos(-1);
    
    int n,m,bit,len=1;
    int rev[N];
    
    struct Complex
    {
        double x,y;
        Complex(double xx=0,double yy=0){x=xx,y=yy;}
        Complex operator + (const Complex &A){return Complex(x+A.x,y+A.y);}
        Complex operator - (const Complex &A){return Complex(x-A.x,y-A.y);}
        Complex operator * (const Complex &A){return Complex(x*A.x-y*A.y,x*A.y+y*A.x);}
    }a[N],b[N];
    
    void fft(Complex *A,int dft)
    {
        for(int i=0;i<len;++i)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int step=1;step<len;step<<=1)
        {
            Complex wn(cos(pi/step),dft*sin(pi/step));
            for(int j=0;j<len;j+=(step<<1))
            {
                Complex wnk(1,0);
                for(int k=j;k<step+j;++k)
                {
                    Complex x=A[k],y=wnk*A[k+step];
                    A[k]=x+y,A[k+step]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==1) return;
        for(int i=0;i<=n+m;++i) A[i].x/=len;
    }
    
    int main()
    {
        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();
        while(len<=n+m) len<<=1,++bit;
        for(int i=0;i<len;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
        fft(a,1),fft(b,1);
        for(int i=0;i<=len;++i) a[i]=a[i]*b[i];
        fft(a,-1);
        for(int i=0;i<=n+m;++i)
            cout<<(int)(a[i].x+0.5)<<' ';
        return 0;
    }
    View Code

    P3338 [ZJOI2014]力

    不会,抄的

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    typedef long long LL;
    
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
    
    const int N=4e5+5;
    const double pi=acos(-1);
    
    int n,m,bit,len=1,rev[N];
    
    struct Complex
    {
        double x,y;
        Complex(double xx=0,double yy=0){x=xx,y=yy;}
        Complex operator + (const Complex &A){return Complex(x+A.x,y+A.y);}
        Complex operator - (const Complex &A){return Complex(x-A.x,y-A.y);}
        Complex operator * (const Complex &A){return Complex(x*A.x-y*A.y,x*A.y+y*A.x);}
    }a[N],b[N],c[N];
    
    void fft(Complex *A,int dft)
    {
        for(int i=0;i<len;++i)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int i=1;i<len;i<<=1)
        {
            Complex wn(cos(pi/i),dft*sin(pi/i));
            for(int j=0,t=i*2;j<len;j+=t)
            {
                Complex wnk(1,0);
                for(int k=j;k<i+j;++k)
                {
                    Complex x=A[k],y=wnk*A[k+i];
                    A[k]=x+y,A[k+i]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==1) return;
        for(int i=0;i<=len;++i) A[i].x/=len;
    }
    
    int main()
    {
        n=read();
        for(int i=1;i<=n;++i) scanf("%lf",&a[i].x),b[n-i+1].x=a[i].x;
        for(int i=1;i<=n;++i) c[i].x=1.0/(1ll*i*i);
        while(len<=(n<<1)) ++bit,len<<=1;
        for(int i=0;i<len;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
        fft(a,1),fft(b,1),fft(c,1);
        for(int i=0;i<=len;++i) a[i]=a[i]*c[i],b[i]=b[i]*c[i];
        fft(a,-1),fft(b,-1);
        for(int i=1;i<=n;++i) printf("%.3lf
    ",a[i].x-b[n-i+1].x);
        return 0;
    }
    View Code

    2194: 快速傅立叶之二

    不会,抄的。 好像知道fft/卷积一拉子是干什么的了。

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
     
    typedef long long LL;
     
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
     
    const int N=4e5+5;
    const double pi=acos(-1);
     
    int n,bit,len=1,rev[N];
     
    struct Complex
    {
        double x,y;
        Complex(double a=0,double b=0){x=a,y=b;}
        Complex operator + (const Complex &A){return Complex(x+A.x,y+A.y);}
        Complex operator - (const Complex &A){return Complex(x-A.x,y-A.y);}
        Complex operator * (const Complex &A){return Complex(x*A.x-y*A.y,x*A.y+y*A.x);}
    }a[N],b[N];
     
    void fft(Complex *A,int dft)
    {
        for(int i=0;i<len;++i)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int i=1;i<len;i<<=1)
        {
            Complex wn(cos(pi/i),dft*sin(pi/i));
            for(int j=0,t=i<<1;j<len;j+=t)
            {
                Complex wnk(1,0);
                for(int k=j;k<i+j;++k)
                {
                    Complex x=A[k],y=wnk*A[k+i];
                    A[k]=x+y,A[k+i]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==1) return;
        for(int i=0;i<len;++i) A[i].x/=len;
    }
     
    int main()
    {
        n=read();
        for(int i=0;i<n;++i)
        {
            a[n-i-1].x=read();
            b[i].x=read();
        }
        while(len<=(n<<1)) len<<=1,++bit;
        for(int i=0;i<len;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
        fft(a,1),fft(b,1);
        for(int i=0;i<=len;++i) a[i]=a[i]*b[i];
        fft(a,-1);
        for(int i=0;i<n;++i)
            printf("%d
    ",(int)(a[n-i-1].x+0.5));
        return 0;
    }
    View Code

    3513: [MUTC2013]idiots

    不会,抄的。还是不知道fft/卷积一拉子是干什么的。

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
     
    typedef long long LL;
     
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
     
    const int N=4e5+5;
    const double pi=acos(-1);
     
    int n,bit,len=1,rev[N];
    LL g[N],t[N];
    
    struct Complex
    {
        double x,y;
        Complex(double a=0,double b=0){x=a,y=b;}
        Complex operator + (const Complex &A){return Complex(x+A.x,y+A.y);}
        Complex operator - (const Complex &A){return Complex(x-A.x,y-A.y);}
        Complex operator * (const Complex &A){return Complex(x*A.x-y*A.y,x*A.y+y*A.x);}
    }f[N];
     
    void fft(Complex *A,int dft)
    {
        for(int i=0;i<len;++i)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int i=1;i<len;i<<=1)
        {
            Complex wn(cos(pi/i),dft*sin(pi/i));
            for(int j=0,t=i<<1;j<len;j+=t)
            {
                Complex wnk(1,0);
                for(int k=j;k<i+j;++k)
                {
                    Complex x=A[k],y=wnk*A[k+i];
                    A[k]=x+y,A[k+i]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==1) return;
        for(int i=0;i<=len;++i) A[i].x/=len;
    }
     
    int main()
    {
        int T=read();
        while(T--)
        {
            memset(g,0,sizeof(g)),memset(t,0,sizeof(t));
            memset(f,0,sizeof(f));
            n=read();int mx=0;LL sum;
            for(int i=1,a;i<=n;++i)
            {
                a=read();mx=max(mx,a);
                --g[a<<1],++t[a],++f[a].x;
            }
            for(int i=mx;i;--i) t[i]+=t[i+1];
            mx<<=1,sum=1ll*n*(n-1)*(n-2)/6;
            len=1,bit=0;while(len<=mx) len<<=1,++bit;
            for(int i=0;i<=len;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
            fft(f,1);
            for(int i=0;i<=len;++i) f[i]=f[i]*f[i];
            fft(f,-1);
            for(int i=0;i<=len;++i) g[i]+=1ll*(f[i].x+0.5);
            LL ans=0;
            for(int i=0;i<=len;++i) ans+=(g[i]/2)*t[i];
            ans=sum-ans;
            printf("%.7lf
    ",1.0*ans/sum);
        }
        return 0;
    }
    View Code
     不会,抄的。
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    typedef long long LL;
    
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
    
    const int N=3e5+5;
    const double pi=acos(-1);
    
    int n,m,len=1,bit,rev[N];
    
    struct Complex
    {
        double x,y;
        Complex(double a=0,double b=0){x=a,y=b;}
        Complex operator + (const Complex &A){return Complex(x+A.x,y+A.y);}
        Complex operator - (const Complex &A){return Complex(x-A.x,y-A.y);}
        Complex operator * (const Complex &A){return Complex(x*A.x-y*A.y,x*A.y+y*A.x);}
    }a[N],b[N];
    
    void fft(Complex *A,int dft)
    {
        for(int i=0;i<len;++i)
            if(i<rev[i]) swap(A[i],A[rev[i]]);
        for(int i=1;i<len;i<<=1)
        {
            Complex wn(cos(pi/i),dft*sin(pi/i));
            for(int j=0,t=i<<1;j<len;j+=t)
            {
                Complex wnk(1,0);
                for(int k=j;k<i+j;++k)
                {
                    Complex x=A[k],y=wnk*A[k+i];
                    A[k]=x+y,A[k+i]=x-y;
                    wnk=wnk*wn;
                }
            }
        }
        if(dft==1) return;
        for(int i=0;i<=len;++i) A[i].x=A[i].x/len+0.5;
    }
    
    int main()
    {
        n=read(),m=read();LL a1=0,a2=0,b1=0,b2=0;
        for(int i=1;i<=n;++i)
        {
            a[i].x=a[n+i].x=read();
            a1+=a[i].x*a[i].x,a2+=a[i].x;
        }
        for(int i=1;i<=n;++i)
        {
            b[n-i+1].x=read();
            b1+=b[n-i+1].x*b[n-i+1].x,b2+=b[n-i+1].x;
        }
        while(len<=(n*3)) len<<=1,++bit;
        for(int i=0;i<len;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
        fft(a,1),fft(b,1);
        for(int i=0;i<=len;++i) a[i]=a[i]*b[i];
        fft(a,-1);
        LL ans=1ll<<60,sum=ans;
        for(int j=-m;j<=m;++j) sum=min(sum,a1+b1+n*j*j+2ll*j*(a2-b2));
        for(int i=1;i<=n;++i) ans=min(ans,sum-2ll*(LL)a[n+i].x);
        cout<<ans;
        return 0;
    }
    View Code

    真tm难啊

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  • 原文地址:https://www.cnblogs.com/lovewhy/p/10217235.html
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