[BZOJ2194] 快速傅里叶之二 题解

题意：求\[C_k=\sum_{k}^{n-1}a_ib_{i-k}.\]
\(n \leq 1e5\).
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考虑反转数组\(a\)，生成新数组\(a'\).
那么\[C_k=\sum_{i=k}^{n-1}a'_{n-1-i}b_{i-k},\]
考虑把\(i\)改成从\(0\)开始。那么
\[C_k=\sum_{i=0}^{n-k-1}a'_{n-k-1-i}b_i.\]
考虑用\(\text{FFT}\)计算卷积的标准形式：
\[C'_x=\sum_{i=0}^{x}A_{x-i}B_i.\]
考虑\(x=n-k-1\)时的情况：
\[C'_{n-k-1}=\sum_{i=0}^{n-k-1}a'_{n-k-1-i}b_i.\]
发现\(C'\)本质上就是将\(C\)的前\(n\)个反转了一下，而\(C'\)是可以直接计算的。
于是，可以直接用\(\text{FFT}\)计算出\(a'\)和\(b\)的卷积\(C'\)，再反转一下前\(n\)项，输出前\(n\)项即可。

#include <bits/stdc++.h>
#define REP(i , x , y) for(__typeof(y) i = x; i <= y; i++)
#define PER(i , y , x) for(__typeof(x) i = y; i >= x; i--)
typedef long long LL;
typedef long double ld;
typedef unsigned long long ULL;
/* do not give up ! try your best! Read the meaning clearly! */
template < typename T > void Input(T& t) {
    char c = getchar(); T x = 1; t = 0; while(!isdigit(c)) {if(c == '-') x = -1; c = getchar();}
    while(isdigit(c)) t = t * 10 + c - '0' , c = getchar();t *= x;
}
template < typename T , typename... Args > void Input(T& t , Args&... args) {Input(t); Input(args...);}
template < typename T > T mul(T x , T y , T _) {x %= _,y %= _;return ((x * y - (T)(((ld)x * y + 0.5) / _) * _) % _ + _) % _;}

using namespace std;

const int MAXN = 2097152 + 10;
const double Pi = acos(-1.0);
int a[MAXN] , b[MAXN] , c[MAXN] , n;
static complex < double > A[MAXN] , B[MAXN] , C[MAXN];
struct FastFourierTransform {
    complex < double > omega[MAXN] , omegaInverse[MAXN];
    void init(int n) {
        for(int i = 0; i < n; i++) {
            omega[i] = complex < double > (cos(2 * Pi / n * i) , sin(2 * Pi / n * i));
            omegaInverse[i] = conj(omega[i]);
        }
    }
    void Transform(complex < double > *a , const int n , const complex < double > *omega) {
        int k = 1;
        while((1 << k) < n) ++k;
        for(int i = 0; i < n; i++) {
            int t = 0;
            for(int j = 0;j < k; j++)
                if(i & (1 << j)) t |= (1 << (k - 1 - j));
            if(i < t) swap(a[i] , a[t]);
        }

        for(int l = 2; l <= n; l <<= 1) {
            int m = (l >> 1);
            for(complex < double > *p = a; p != a + n; p += l) {
                for(int i = 0; i < m; i++) {
                    complex < double > k = omega[n / l * i] * p[m + i];
                    p[m + i] = p[i] - k;
                    p[i] += k;
                }
            }
        }
    }
    void DFT(complex < double > *a , int n) {
        Transform(a , n , omega);
    }
    void IDFT(complex < double > *a , int n) {
        Transform(a , n , omegaInverse);
        for(int i = 0; i < n; i++) a[i] /= n;
    }
}FFT;
int main() {
    Input(n);
    for(int i = 0; i < n; i++) Input(a[i] , b[i]);
    reverse(a , a + n);
    for(int i = 0; i < n; i++) A[i].real(a[i]) , B[i].real(b[i]);
    int N = 1;
    while(N < 2 * n) N <<= 1;
    FFT.init(N);
    FFT.DFT(A , N); FFT.DFT(B , N);
    for(int i = 0; i < N; i++) C[i] = A[i] * B[i];
    FFT.IDFT(C , N);
    for(int i = 0; i < N; i++) c[i] = static_cast < int > (C[i].real() + 0.5);
    int cnt = 0;
    for(int i = 0; i < n; i++) printf("%d\n" , c[n - 1 - i]);
    return 0;
}

标签：int,题解,void,BZOJ2194,++,complex,MAXN,double,傅里叶
来源： https://www.cnblogs.com/LiM-817/p/10887171.html