題目鏈接

BZOJ4503

題解

水水題。
和殘缺的字符串那題幾乎是一樣的
同樣轉化為多項式
同樣TLE
同樣要手寫一下復數才A

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<complex>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
 

#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if
 
 (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
struct E{
    double a,b;
    E(){}
    E(double x,double y):a(x),b(y) {}
    E(int x,int y):a(x),b(y) {}
    inline E operator
 
 =(const int& b){
        this->a = b; this->b = 0;
        return *this;
    }
    inline E operator =(const double& b){
        this->a = b; this->b = 0;
        return *this;
    }
    inline E operator /=(const double& b){
        this->a /= b; this->b /= b;
        return *this;
    }
};
inline E operator *(const E& a,const E& b){
    return E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator *=(E& a,const E& b){
    return a = E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator +(const E& a,const E& b){
    return E(a.a + b.a,a.b + b.b);
}
inline E operator -(const E& a,const E& b){
    return E(a.a - b.a,a.b - b.b);
}
const double pi = acos(-1);
int R[maxn];
void fft(E* a,int n,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        E wn(cos(pi / i),f * sin(pi / i));
        for (int j = 0; j < n; j += (i << 1)){
            E w(1,0),x,y;
            for (int k = 0; k < i; k++,w = w * wn){
                x = a[j + k],y = w * a[j + k + i];
                a[j + k] = x + y; a[j + k + i] = x - y;
            }
        }
    }
    if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
E A[maxn],B[maxn];
int N,M,L,ans[maxn],ansi;
double C[maxn];
char S[maxn],T[maxn];
int main(){
    scanf("%s",S); N = strlen(S);
    scanf("%s",T); M = strlen(T);
    reverse(T,T + M);
    int n,m; double t;
    m = N - 1 + M - 1; L = 0;
    for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i < N; i++){
        t = S[i] - ‘a‘ + 1;
        A[i] = t * t;
    }
    for (int i = 0; i < M; i++)
        if (T[i] == ‘?‘) B[i] = 0;
        else B[i] = T[i] - ‘a‘ + 1;
    fft(A,n,1); fft(B,n,1);
    for (int i = 0; i < n; i++) A[i] *= B[i];
    fft(A,n,-1);
    for (int i = 0; i < N; i++) C[i] += floor(A[i].a + 0.5);
        
    for (int i = 0; i < N; i++) A[i] = 1;
    for (int i = N; i < n; i++) A[i] = 0;
    for (int i = 0; i < M; i++)
        if (T[i] == ‘?‘) B[i] = 0;
        else {
            t = T[i] - ‘a‘ + 1;
            B[i] = t * t * t;
        }
    for (int i = M; i < n; i++) B[i] = 0;
    fft(A,n,1); fft(B,n,1);
    for (int i = 0; i < n; i++) A[i] *= B[i];
    fft(A,n,-1);
    for (int i = 0; i < N; i++) C[i] += floor(A[i].a + 0.5);
    
    for (int i = 0; i < N; i++)
        A[i] = S[i] - ‘a‘ + 1;
    for (int i = N; i < n; i++) A[i] = 0;
    for (int i = 0; i < M; i++)
        if (T[i] == ‘?‘) B[i] = 0;
        else {
            t = T[i] - ‘a‘ + 1;
            B[i] = t * t;
        }
    for (int i = M; i < n; i++) B[i] = 0;
    fft(A,n,1); fft(B,n,1);
    for (int i = 0; i < n; i++) A[i] *= B[i];
    fft(A,n,-1);
    for (int i = 0; i < N; i++) C[i] -= 2 * floor(A[i].a + 0.5);
    for (int i = M - 1; i < N; i++)
        if (fabs(C[i]) < 0.1) ans[++ansi] = i - M + 1;
    printf("%d\n",ansi);
    REP(i,ansi) printf("%d\n",ans[i]);
    return 0;
}

BZOJ4503 兩個串 【fft】