bzoj 4503 两个串 FFT
题面
解法
挺妙的一道题，思路比较精妙
带通配符之后好像字符串算法就不太好匹配了啊
设通配符的值为0，那么这两个字符相同当且仅当有一个是通配符或者就是相同的
设\(dis(A,B)=\sum_{i=0}^{n-1}(A_i-B_i)^2A_iB_i\)
其中，若\(A_i=?\)，那么\(A_i=0\)
如果\(dis(A,B)=0\)，那么显然这两个字符串是匹配的
\(p_i\)表示\(B\)以\(i\)结尾的字符串和\(A\)的匹配情况
翻转A串，然后可以得出以下式子：
\[p_i=\sum_{j=0}^{n-1}A_j^3B_{i-j}-2\sum_{j=0}^{n-1}A_j^2B_{i-j}^2+\sum_{j=0}^{n-1}A_jB_{i-j}^3$
显然，这是一个卷积的形式，可以FFT
分三次FFT即可
时间复杂度：$O(n\ log \ n)$
~~常数巨大~~
### 代码
```cpp
#include <bits/stdc++.h>
#define N 1 << 21
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Complex {
double x, y;
Complex (double tx = 0, double ty = 0) {x = tx, y = ty;}
} a[N], b[N], c[N];
const double pi = acos(-1);
int n, m, A[N], B[N], ans[N], rev[N];
Complex operator + (Complex a, Complex b) {return (Complex) {a.x + b.x, a.y + b.y};}
Complex operator - (Complex a, Complex b) {return (Complex) {a.x - b.x, a.y - b.y};}
Complex operator * (Complex a, Complex b) {return (Complex) {a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
Complex operator * (Complex a, double b) {return (Complex) {a.x * b, a.y * b};}
string tx, ty;
void getrev(int l) {
for (int i = 0; i < (1 << l); i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << l - 1);
}
void FFT(Complex *a, int n, int key) {
for (int i = 0; i < n; i++)
if (i < rev[i]) swap(a[i], a[rev[i]]);
for (int i = 1; i < n; i <<= 1) {
Complex wn(cos(pi / i), key * sin(pi / i));
for (int r = i << 1, j = 0; j < n; j += r) {
Complex w(1, 0);
for (int k = 0; k < i; k++, w = w * wn) {
Complex x = a[j + k], y = w * a[i + j + k];
a[j + k] = x + y, a[i + j + k] = x - y;
}
}
}
if (key == -1)
for (int i = 0; i < n; i++) a[i].x /= n;
}
int main() {
cin >> tx >> ty;
int m = tx.size(), n = ty.size();
for (int i = 0; i < n; i++)
if (ty[i] != '?') A[i] = ty[i] - 'a' + 1;
for (int i = 0; i < m; i++)
if (tx[i] != '?') B[i] = tx[i] - 'a' + 1;
int len = 1, l = 0;
while (len <= 2 * m) len <<= 1, l++; getrev(l); reverse(A, A + n);
for (int i = 0; i < len; i++) a[i] = (Complex) {A[i] * A[i] * A[i], 0};
for (int i = 0; i < len; i++) b[i] = (Complex) {B[i], 0};
FFT(a, len, 1), FFT(b, len, 1);
for (int i = 0; i < len; i++) c[i] = c[i] + (a[i] * b[i]);
for (int i = 0; i < len; i++) a[i] = (Complex) {A[i] * A[i], 0};
for (int i = 0; i < len; i++) b[i] = (Complex) {B[i] * B[i], 0};
FFT(a, len, 1), FFT(b, len, 1);
for (int i = 0; i < len; i++) c[i] = c[i] - (a[i] * b[i] * 2.0);
for (int i = 0; i < len; i++) a[i] = (Complex) {A[i], 0};
for (int i = 0; i < len; i++) b[i] = (Complex) {B[i] * B[i] * B[i], 0};
FFT(a, len, 1), FFT(b, len, 1);
for (int i = 0; i < len; i++) c[i] = c[i] + (a[i] * b[i]);
FFT(c, len, -1); int tot = 0;
for (int i = n - 1; i < m; i++)
if (fabs(c[i].x) < 0.5) ans[++tot] = i - n + 1;
cout << tot << "\n";
for (int i = 1; i <= tot; i++) cout << ans[i] << ' ';
return 0;
}
```\]
