• 多项式乘法(代码)


    多项式乘法

    模板题, 下面给出一个很方便使用的模板,可当做黑盒代码使用。

    #include <set>
    #include <cmath>
    #include <cctype>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <assert.h>
    #include <algorithm>
    
    using namespace std;
    
    #define fir first
    #define sec second
    #define pb push_back
    #define mp make_pair
    #define LL long long
    #define INF (0x3f3f3f3f)
    #define mem(a, b) memset(a, b, sizeof (a))
    #define debug(...) fprintf(stderr, __VA_ARGS__)
    #define Debug(x) cout << #x << " = " << x << endl
    #define travle(i, x) for (register int i = head[x]; i; i = nxt[i])
    #define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
    #define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
    #define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
    #define ____ debug("go
    ")
    
    namespace io {
        static char buf[1<<21], *pos = buf, *end = buf;
        inline char getc()
        { return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
        inline int rint() {
            register int x = 0, f = 1;register char c;
            while (!isdigit(c = getc())) if (c == '-') f = -1;
            while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
            return x * f;
        }
        inline LL rLL() {
            register LL x = 0, f = 1; register char c;
            while (!isdigit(c = getc())) if (c == '-') f = -1;
            while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
            return x * f;
        }
        inline void rstr(char *str) {
            while (isspace(*str = getc()));
            while (!isspace(*++str = getc()))
                if (*str == EOF) break;
            *str = '';
        }
        template<typename T> 
            inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
        template<typename T>
            inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }    
    }
    using namespace io;
    
    
    const int N = 4e6 + 10;
    const double PI = acos(-1);
    struct vec 
    {
        double x, y;
        vec(double xx = 0, double yy = 0) { x = xx, y = yy; }
        friend vec operator+ (const vec &a, const vec &b)
        { return vec(a.x + b.x, a.y + b.y); }
        friend vec operator- (const vec &a, const vec &b)
        { return vec(a.x - b.x, a.y - b.y); }
        friend vec operator* (const vec &a, const vec &b)
        { return vec(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); }
    };
    vec omega(int n, int k)
    { return vec(cos(2 * PI / n * k), sin(2 * PI / n * k)); }
    
    namespace FFT 
    {
        int r[N];
        inline void DFT(vec *f, int type, int lim)
        {
            for (int i = 0; i < lim; ++ i)
                if (i < r[i])
                    swap(f[i], f[r[i]]);
            for (int l = 2; l <= lim; l <<= 1) 
            {
                int k = l >> 1;
                for (vec *p = f; p != f + lim; p += l)
                    for (int i = 0; i < k; ++ i)
                    {
                        vec t = omega(l, type * i) * p[i + k];
                        p[i + k] = p[i] - t;
                        p[i] = p[i] + t;
                    }
            }
        }
        inline void Mul(double *a, int n1, double *b, int n2, double *ans) 
        {
            int lim = 1, l = 0;
            while (lim <= n1 + n2) lim <<= 1, l++;
            for (int i = 0; i < lim; ++ i)
                r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
            static vec a1[N], b1[N];
            for (int i = 0; i < lim; ++ i) 
            {
                a1[i] = vec(i <= n1 ? a[i] : 0, 0);
                b1[i] = vec(i <= n2 ? b[i] : 0, 0);
            }
            DFT(a1, 1, lim); DFT(b1, 1, lim);
            for (int i = 0; i <= lim; ++ i)
                a1[i] = a1[i] * b1[i];
            DFT(a1, -1, lim);
            for (int i = 0; i <= n1 + n2; ++ i)
                ans[i] = a1[i].x / lim;
        }
    }
    int main() {
    #ifndef ONLINE_JUDGE
        file("FFT_Template");
    #endif
        int n, m;
        n = rint(), m = rint();
        static double a[N], b[N], c[N];
        for (int i = 0; i <= n; ++ i) a[i] = rint();
        for (int i = 0; i <= m; ++ i) b[i] = rint();
        FFT::Mul(a, n, b, m, c);
        for (int i = 0; i <= n + m; ++ i)
            printf("%d ", (int)(c[i] + 0.5));
    }
    
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  • 原文地址:https://www.cnblogs.com/cnyali-Tea/p/10539970.html
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