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  • FFT笔记

    机房的人都嘲笑我博客只有两句话,我今天就要打他们的脸!!!我今天写一下FFT这个神奇的算法.首先,我们需要知道,FFT好像就是用来处理两个多项式乘积的,没有多么高端.首先思考暴力,就是暴力把n个数带入求值,然后相乘,复杂度O(n^2).我们考虑分治,分治奇偶,把奇数提出来,再进行分治.先放一个递归的简单易懂的代码:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    #include<vector>
    #include <complex>
    #include<cstring>
    using namespace std;
    #define duke(i,a,n) for(int i = a;i <= n;i++)
    #define lv(i,a,n) for(int i = a;i >= n;i--)
    #define clean(a) memset(a,0,sizeof(a))
    #define mp make_pair
    #define cp complex<db>
    #define enter puts("")
    const int INF = 1 << 30;
    const double eps = 1e-8;
    typedef long long ll;
    typedef double db;
    template <class T>
    void read(T &x)
    {
        char c;
        bool op = 0;
        while(c = getchar(), c < '0' || c > '9')
            if(c == '-') op = 1;
        x = c - '0';
        while(c = getchar(), c >= '0' && c <= '9')
            x = x * 10 + c - '0';
        if(op) x = -x;
    }
    template <class T>
    void write(T x)
    {
        if(x < 0) putchar('-'), x = -x;
        if(x >= 10) write(x / 10);
        putchar('0' + x % 10);
    }
    const int N = 4000005;
    const db PI = acos(-1);
    cp a[N],b[N],omg[N],inv[N];
    int len = 1,ans[N];
    bool inversed = false;
    cp omega(int n,int k)
    {
        if(inversed == true)
        return cp(cos(2 * PI * k / n),sin(2 * PI * k / n));
        else
        return conj(cp(cos(2 * PI * k / n),sin(2 * PI * k / n)));
    }
    void fft(cp *a,int n)
    {
        if(n == 1) return;
        static cp buf[N];
        int m = n / 2;
        duke(i,0,m - 1)
        {
            buf[i] = a[2 * i];
            buf[i + m] = a[2 * i + 1];
        }
        for(int i = 0;i < n;i++)
        {
            a[i] = buf[i];
        }
        fft(a,m);
        fft(a + m,m);
        duke(i,0,m - 1)
        {
            cp x = omega(n,i);
            buf[i] = a[i] + x * a[i + m];
            buf[i + m] = a[i] - x * a[i + m];
        }
        duke(i,0,n - 1)
        a[i] = buf[i];
    }
    int n,m;
    int main()
    {
        read(n);read(m);
        while(len < n + m + 1) len <<= 1;
        // cout<<len<<endl;
        int t;
        duke(i,0,n)
        {
            read(t);
            a[i].real(t);
        }
        duke(i,0,m)
        {
            read(t);
            b[i].real(t);
        }
        fft(a,len);fft(b,len);
        duke(i,0,len)
        {
            a[i] *= b[i];
        }
        inversed = true;
        fft(a,len);
        duke(i,0,len)
        {
            ans[i] = floor(a[i].real() / len + 0.5);
        }
        duke(i,0,n + m)
        {
            printf("%d ",ans[i]);
        }
        return 0;
    }

    这个代码简单易懂,但是luogu会T2个点,怎么办呢,我们尝试2进制优化,每次直接处理出现在这位的数,得到非递归版本的FFT,上一下代码:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    #include<vector>
    #include <complex>
    #include<cstring>
    using namespace std;
    #define duke(i,a,n) for(int i = a;i <= n;i++)
    #define lv(i,a,n) for(int i = a;i >= n;i--)
    #define clean(a) memset(a,0,sizeof(a))
    #define mp make_pair
    #define cp complex<db>
    #define enter puts("")
    const int INF = 1 << 30;
    const double eps = 1e-8;
    typedef long long ll;
    typedef double db;
    template <class T>
    void read(T &x)
    {
        char c;
        bool op = 0;
        while(c = getchar(), c < '0' || c > '9')
            if(c == '-') op = 1;
        x = c - '0';
        while(c = getchar(), c >= '0' && c <= '9')
            x = x * 10 + c - '0';
        if(op) x = -x;
    }
    template <class T>
    void write(T x)
    {
        if(x < 0) putchar('-'), x = -x;
        if(x >= 10) write(x / 10);
        putchar('0' + x % 10);
    }
    const int N = 4000005;
    const db PI = acos(-1);
    cp a[N],b[N],omg[N],inv[N];
    int len = 1,ans[N];
    void init(int n)
    {
        duke(i,0,n - 1)
        {
            omg[i] = cp(cos(2 * PI * i / n),sin(2 * PI * i / n));
            inv[i] = conj(omg[i]);
        }
    }
    void fft(cp *a,int n)
    {
        int lim = 0;
        while((1 << lim) < n) lim++;
        duke(i,0,n - 1)
        {
            int t = 0;
            duke(j,0,lim - 1)
            {
                if((i >> j) & 1) t |= (1 << (lim - j - 1));
            }
            if(i < t) swap(a[i],a[t]);
        }
        static cp buf[N];
        for(int l = 2;l <= n;l *= 2)
        {
            int m = l / 2;
            for(int j = 0;j < n;j += l)
            {
                for(int i = 0;i < m;i++)
                {
                    // cp x = omega(n / l,i);
                    buf[j + i] = a[j + i] + omg[n / l * i] * a[j + i + m];
                    buf[j + i + m] = a[j + i] - omg[n / l * i] * a[j + i + m];
                }
            }
            duke(i,0,n - 1)
            a[i] = buf[i];
        }
    }
    int n,m;
    int main()
    {
        read(n);read(m);
        while(len < n + m + 1) len <<= 1;
        // cout<<len<<endl;
        int t;
        duke(i,0,n)
        {
            read(t);
            a[i].real(t);
        }
        duke(i,0,m)
        {
            read(t);
            b[i].real(t);
        }
        init(len);
        fft(a,len);fft(b,len);
        duke(i,0,len)
        {
            a[i] *= b[i];
        }
        duke(i,0,len)
        {
            omg[i] = inv[i];
        }
        fft(a,len);
        duke(i,0,len)
        {
            ans[i] = floor(a[i].real() / len + 0.5);
        }
        duke(i,0,n + m)
        {
            printf("%d ",ans[i]);
        }
        return 0;
    }

    但还是很慢,还是会T两个点,我们需要用一个叫蝴蝶优化的东西,就是把操作顺序改变一下就行了.具体其实我不太懂,先背板子以后慢慢理解吧:

    // luogu-judger-enable-o2
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    #include<vector>
    #include <complex>
    #include<cstring>
    using namespace std;
    #define duke(i,a,n) for(int i = a;i <= n;i++)
    #define lv(i,a,n) for(int i = a;i >= n;i--)
    #define clean(a) memset(a,0,sizeof(a))
    #define mp make_pair
    #define cp complex<db>
    #define enter puts("")
    const int INF = 1 << 30;
    const double eps = 1e-8;
    typedef long long ll;
    typedef double db;
    template <class T>
    void read(T &x)
    {
        char c;
        bool op = 0;
        while(c = getchar(), c < '0' || c > '9')
            if(c == '-') op = 1;
        x = c - '0';
        while(c = getchar(), c >= '0' && c <= '9')
            x = x * 10 + c - '0';
        if(op) x = -x;
    }
    template <class T>
    void write(T x)
    {
        if(x < 0) putchar('-'), x = -x;
        if(x >= 10) write(x / 10);
        putchar('0' + x % 10);
    }
    const int N = 4000005;
    const db PI = acos(-1);
    cp a[N],b[N],omg[N],inv[N];
    int len = 1,ans[N];
    void init(int n)
    {
        duke(i,0,n - 1)
        {
            omg[i] = cp(cos(2 * PI * i / n),sin(2 * PI * i / n));
            inv[i] = conj(omg[i]);
        }
    }
    void fft(cp *a,int n)
    {
        int lim = 0;
        while((1 << lim) < n) lim++;
        duke(i,0,n - 1)
        {
            int t = 0;
            duke(j,0,lim - 1)
            {
                if((i >> j) & 1) t |= (1 << (lim - j - 1));
            }
            if(i < t) swap(a[i],a[t]);
        }
        for(int l = 2;l <= n;l *= 2)
        {
            int m = l / 2;
            for(cp *p = a;p != a + n;p += l)
            {
                for(int i = 0;i < m;i++)
                {
                    cp t = omg[n / l * i] * p[i + m];
                    p[i + m] = p[i] - t;
                    p[i] += t;
                }
            }
        }
    }
    int n,m;
    int main()
    {
        read(n);read(m);
        while(len < n + m + 1) len <<= 1;
        // cout<<len<<endl;
        int t;
        duke(i,0,n)
        {
            read(t);
            a[i].real(t);
        }
        duke(i,0,m)
        {
            read(t);
            b[i].real(t);
        }
        init(len);
        fft(a,len);fft(b,len);
        duke(i,0,len)
        {
            a[i] *= b[i];
        }
        duke(i,0,len)
        {
            omg[i] = inv[i];
        }
        fft(a,len);
        duke(i,0,len)
        {
            ans[i] = floor(a[i].real() / len + 0.5);
        }
        duke(i,0,n + m)
        {
            printf("%d ",ans[i]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/DukeLv/p/10064171.html
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