BZOJ2194 快速傅立叶之二

于是标题又暴露了一切

把b倒过来读进来，就变成了卷积（那是什么。。。？可以吃吗？）的形式

于是直接FFT一下就好了

/**************************************************************
    Problem: 2194
    User: rausen
    Language: C++
    Result: Accepted
    Time:2760 ms
    Memory:21900 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <algorithm>
 
#define complex P
using namespace std;
typedef double lf;
const int N = 300005;
const lf pi = acos(-1);
 
struct complex {
    lf x, y;
    P() {}
    P(lf _x, lf _y) : x(_x), y(_y) {}
     
    inline P operator + (const P &X) const {
        return P(x + X.x, y + X.y);
    }
    inline P operator - (const P &X) const {
        return P(x - X.x, y - X.y);
    }
    inline P operator * (const P &X) const {
        return P(x * X.x - y * X.y, x * X.y + y * X.x);
    }
     
    inline void operator += (const P &X) {
        *this = *this + X;
    }
    inline void operator *= (const P &X) {
        *this = *this * X;
    }
} x[N], y[N], w[2][N];
 
int n, len;
int a[N], b[N];
 
inline int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
void FFT(P *x, int len, int v) {
    int i, j, l;
    P tmp;
    for (i = j = 0; i < len; ++i) {
        if (i > j) swap(x[i], x[j]);
        for (l = len >> 1; (j ^= l) < l; l >>= 1);
    }
    for (i = 2; i <= len; i <<= 1)
        for (j = 0; j < len; j += i)
            for (l = 0; l < i >> 1; ++l) {
                tmp = x[j + l + (i >> 1)] * w[v][len / i * l];
                x[j + l + (i >> 1)] = x[j + l] - tmp;
                x[j + l] += tmp;
            }
}
 
void work() {
    int i;
    for (len = 1; len < n << 1; len <<= 1);
    for (i = 0; i <= len; ++i)
        w[1][len -  i] = w[0][i] = P(cos(pi * 2 * i / len), sin(pi * 2 * i / len));
         
    for (i = 0; i < len; ++i)
        x[i] = P(a[i], 0), y[i] = P(b[i], 0);
    FFT(x, len, 0), FFT(y, len, 0);
     
    for (i = 0; i < len; ++i)
        x[i] *= y[i];
    FFT(x, len, 1);
 
    for (i = n - 1; i + 1 < n << 1; ++i)
        printf("%d
", (int) round(x[i].x / len));
}
 
int main() {
    int i;
    n = read();
    for (i = 0; i < n; ++i)
        a[i] = read(), b[n - i - 1] = read();
    work();
    return 0;
}

（p.s.  Xs我回来啦！！！）