gym101002K. Inversions (FFT)

题意:给定一个仅含有AB的字母串
　　 如果i有一个B j有一个A 且j>i 会对F(j-i)产生贡献 求出所有发Fi
题解:好像是很裸的FFT B的分布可以看作一个多项式 同理A也可以
　　 然后把B的位置翻转一下 就搞成了卷积的形式

　　 设f为B的位置函数 如果si = B, fi = 1否则fi = 0. 设g为A的位置函数

\[F(i)= \sum_{j = 1}^{n - i + 1} f(j)*g(i+j)\]

\[把f翻转一下\]
\[F(i)= \sum_{j = 1}^{n - i + 1} f(n - j + 1)*g(i+j) = F(n + 1 + i)\]

#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);

struct Complex {
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0) {
        x = _x;
        y = _y;
    }
    Complex operator + (const Complex &b) const {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator - (const Complex &b) const {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator * (const Complex &b) const {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};

void change(Complex y[], int len) {
    int i, j, k;
    for(i = 1, j = len / 2; i < len - 1; i++) {
        if(i < j) swap(y[i], y[j]);
        k = len / 2;
        while(j >= k) {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

void fft(Complex y[], int len, int on) {
    change(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for(int j = 0; j < len; j += h) {
            Complex w(1, 0);
            for(int k = j; k < j + h / 2; k++) {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }

    if(on == -1)
        for(int i = 0; i < len; i++)
            y[i].x /= len;
}

char s[1000005];
Complex x1[4000005], x2[4000005];
int main() {
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    for(int i = 1; i <= n; i++) {
        if(s[i] == 'A') {
            x1[i] = Complex(1.0, 0);
            x2[n - i + 1] = Complex(0, 0);
        } else if(s[i] == 'B') {
            x1[i] = Complex(0, 0);
            x2[n - i + 1] = Complex(1.0, 0);
        }
    }
    int len = 1;
    while(len <= n + n) len <<= 1;
    fft(x1, len, 1);
    fft(x2, len, 1);
    for(int i = 0; i <= len; i++) x1[i] = x1[i] * x2[i];
    fft(x1, len, -1);
    for(int i = n + 2; i <= n + n; i++) printf("%d\n", (int)(x1[i].x + 0.5));
    return 0;
}

标签：gym101002K,const,int,double,FFT,len,operator,Complex,Inversions
来源： https://www.cnblogs.com/lwqq3/p/11349099.html