題目連結

題意：給定兩個串A，B，問A中有多少個長度等於B的子串與B相似，兩個同樣長度的字串相似的條件是：對於對應位置上的字元，差的絕對值不大於1

思路： 可以類似於FFT用於字串匹配的演算法。 假設$A$的長度為$m$，$B$的長度為$n$，且下標均從1開始 則相似子串的個數：

$Ans = \sum_{j=0}^{m-1}[\sum_{i=1}^{n}(A_{i+j}-B_i)^2*((A_{i+j}-B_i)^2 - 1) = 0]$

將中間的式子展開： $(A_{i+j}-B_i)^2*((A_{i+j}-B_i)^2 - 1)$

$=(A_{i+j}^2+B_i^2-2A_{i+j}B_i)*(A_{i+j}^2+B_i^2-2A_{i+j}B_i - 1)$ $=(A_{i+j}^4-A_{i+j}^2)+(B_i^4-B_i^2)+6A_i^{}$

前面的四項可以處理一個字首和，而後面的四項可以通過反轉B以後進行卷積，FFT優化。

隨後就可以列舉每個位置，O(1)判斷了

以上。

程式碼：

#include<cstdio>
#include
 
<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const double pi = acos(-1.0);
const int A = 250000 + 10;
const int B = A<<3;
char S[A], T[A], E[30];
int n, m, tot, len, Ans[A];
ll sumx[A], sumy[A];

struct comp{
    double r, i; comp(double _r = 0, double _i = 0) {r = _r; i = _i;}
    comp operator + (const comp x){return comp(r+x.r, i+x.i);}
    comp operator - (const comp x){return comp(r-x.r, i-x.i);}
    comp operator * (const comp x){return comp(r*x.r-i*x.i, r*x.i+i*x.r);}
    comp operator * (const int x){return comp(r*x, i*x);}
}X[4][B], Y[3][B], Z[B];

void FFT(comp a[], int n, int t){
    for (int i = 1, j = 0; i < n - 1; i++) {
        for (int s = n; j ^= s>>=1, ~j&s;);
        if (i < j) swap(a[i], a[j]);
    }
    for (int d = 0; (1<<d) < n; d++) {
        int m = 1<<d, m2 = m<<1;
        double o = pi/m*t; comp _w(cos(o), sin(o));
        for (int i = 0; i < n; i += m2) {
            comp w(1, 0);
            for (int j = 0; j < m; j++) {
                comp &A = a[i+j+m], &B = a[i+j], t = w*A;
                A = B - t; B = B + t; w = w * _w;
            }
        }
    }
    if (t == -1) for(int i = 0; i < n; i++) a[i].r /= n;
}

void Init(){
    len = 1;
    while (len < m+2) len <<= 1;
    sumx[0] = sumy[0] = 0;
    for (int i = 1; i <= n; i++) {
        int x = S[i] - 'a' + 1;
        sumx[i] = sumx[i-1] + (x*x*x*x - x*x);
        X[0][n-i+1] = comp(2.0*x, 0);
    	X[1][n-i+1] = comp(6.0*x*x, 0);
    	X[2][n-i+1] = comp(4.0*x*x*x, 0);
    	X[3][n-i+1] = comp(4.0*x, 0);
    }
    for (int i = 1; i <= m; i++) {
        int y = T[i] - 'a' + 1;
        sumy[i] = sumy[i-1] + (y*y*y*y - y*y);
        Y[0][i] = comp(1.0*y, 0);
    	Y[1][i] = comp(1.0*y*y, 0);
    	Y[2][i] = comp(1.0*y*y*y, 0);
    }
}

void calc(){
    Init();
    for (int i = 0; i < 3; i++) {
        FFT(X[i], len, 1);
        FFT(Y[i], len, 1);
    }
    FFT(X[3], len, 1);
    for (int i = 0; i < len; i++) Z[i] = X[1][i]*Y[1][i] + X[0][i]*Y[0][i] - X[2][i]*Y[0][i] - X[3][i]*Y[2][i];
    FFT(Z, len, -1);

    tot = 0;
    for (int j = 0; j <= m - n; j++) {
        if ((ll)(Z[n+j+1].r + sumx[n] + sumy[n+j] - sumy[j]) == 0) Ans[++tot] = j + 1;
    }
    printf("%d\n", tot);
    for (int i = 1; i <= tot; i++) printf("%d%c", Ans[i], i==tot?'\n':' ');
}

int main(){
    scanf("%s%s%s", S + 1, T + 1, E);
    n = strlen(S + 1);
    m = strlen(T + 1);
    for (int i = 1; i <= n; i++) S[i] = E[S[i]-'a'];
    for (int i = 1; i <= m; i++) T[i] = E[T[i]-'a'];
    calc();
    return 0;
}