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; }

轉載于:https://www.cnblogs.com/Mychael/p/9112828.html

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