FFT NTT模板

FFT与NTT算法详解

最新推荐文章于 2025-12-04 22:00:44 发布

原创最新推荐文章于 2025-12-04 22:00:44 发布 · 185 阅读

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本文深入探讨了快速傅立叶变换（FFT）与数论变换（NTT）的实现原理及应用，提供了完整的C++代码示例。通过具体实例展示了如何使用这两种算法进行多项式乘法。

原题链接https://www.luogu.com.cn/problem/P3803

FFT

#include <bits/stdc++.h>

using ll = long long;
using namespace std;

const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1);

struct com {
    double x, y;

    com(double x1 = 0, double y1 = 0) {
        x = x1, y = y1;
    }
};

com operator+(com a, com b) {
    return com(a.x + b.x, a.y + b.y);
}

com operator-(com a, com b) {
    return com(a.x - b.x, a.y - b.y);
}

com operator*(com a, com b) {
    return com(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}

namespace FFT {
    int tot, digit, rev[maxn << 2];
    com a[maxn << 2], b[maxn << 2];

    void init(int len) {
        tot = 1, digit = 0;
        while (tot <= len) {
            tot <<= 1, digit++;
        }
        for (int i = 0; i < tot; ++i) {
            rev[i] = (rev[i >> 1] >> 1) | (i & 1) << (digit - 1);
        }
    }

    void fft(com *A, int f) {
        for (int i = 0; i < tot; ++i) {
            if (i < rev[i]) swap(A[i], A[rev[i]]);
        }
        for (int mid = 1; mid < tot; mid <<= 1) {
            com w1 = com(cos(pi / mid), f * sin(pi / mid));
            int len = mid * 2;
            for (int p = 0; p < tot; p += len) {
                com wk = com(1, 0);
                for (int k = 0; k < mid; ++k) {
                    com x = A[p + k], y = wk * A[p + k + mid];
                    A[p + k] = x + y;
                    A[p + k + mid] = x - y;
                    wk = wk * w1;
                }
            }
        }
        if (f == -1) {
            for (int i = 0; i < tot; ++i) {
                A[i].x = (int) (A[i].x / tot + 0.5);
            }
        }
    }

    void calc() {
        fft(a, 1);
        fft(b, 1);
        for (int i = 0; i < tot; ++i) {
            a[i] = a[i] * b[i];
        }
        fft(a, -1);
    }

};
using namespace FFT;

int main() {

    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n; ++i) {
        scanf("%lf", &a[i].x);
    }
    for (int i = 0; i <= m; ++i) {
        scanf("%lf", &b[i].x);
    }
    init(n + m);
    calc();
    for (int i = 0; i <= n + m; i++) {
        printf("%.0lf%c", a[i].x, i == n + m ? 10 : 32);
    }
    return 0;
}

NTT

#include <bits/stdc++.h>

using ll = long long;
using namespace std;

const int maxn = 1e6 + 10;

int qmi(int a, int b, int mod) {
    int res = 1;
    while (b) {
        if (b & 1) res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        b >>= 1;
    }
    return res;
}

namespace NTT {
    const int G = 3, P = 998244353, GI = 332748118;
    int tot, digit, rev[maxn << 2];
    int a[maxn << 2], b[maxn << 2];

    void init(int len) {
        tot = 1, digit = 0;
        while (tot <= len) {
            tot <<= 1, digit++;
        }
        for (int i = 0; i < tot; ++i) {
            rev[i] = (rev[i >> 1] >> 1) | (i & 1) << (digit - 1);
//            a[i] = 0, b[i] = 0;
        }
    }

    void ntt(int *A, int f) {
        for (int i = 0; i < tot; ++i) {
            if (i < rev[i]) swap(A[i], A[rev[i]]);
        }
        for (int mid = 1; mid < tot; mid <<= 1) {
            int w1, len = mid * 2;
            if (f == 1) w1 = qmi(G, (P - 1) / len, P);
            else w1 = qmi(GI, (P - 1) / len, P);
            for (int p = 0; p < tot; p += len) {
                int wk = 1;
                for (int k = 0; k < mid; ++k) {
                    int x = A[p + k], y = 1ll * wk * A[p + k + mid] % P;
                    A[p + k] = (1ll * x + y) % P;
                    A[p + k + mid] = (1ll * x - y + P) % P;
                    wk = 1ll * wk * w1 % P;
                }
            }
        }
        if (f == -1) {
            int invp = qmi(tot, P - 2, P);
            for (int i = 0; i < tot; ++i) {
                a[i] = 1ll * a[i] * invp % P;
            }
        }
    }

    void calc() {
        ntt(a, 1), ntt(b, 1);
        for (int i = 0; i < tot; ++i) {
            a[i] = 1ll * a[i] * b[i] % P;
        }
        ntt(a, -1);
    }

};
using namespace NTT;


int main() {

    int n, m;


    scanf("%d%d", &n, &m);
    init(n + m);
    for (int i = 0; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i <= m; ++i) {
        scanf("%d", &b[i]);
    }


    calc();
    for (int i = 0; i <= n + m; i++) {
        printf("%d%c", a[i], i == n + m ? 10 : 32);
    }
    return 0;
}